About the Author
Dr. P K Nag had been with the Indian Institute of Technology, Kharagpur, for about
forty years, almost since his graduation. After retirement from IIT, he was an Emeritus Fellow of AICTE, New Delhi, stationed at Jadavpur University, Kolkata, till June,
2005. He was a Visiting Professor in Technical University of Nova Scotia (now
Dalhousie University), Halifax, Canada, for two years during 1985–86 and 1993–94.
He has authored four books, including this one, and about 150 research papers in
several national and international journals and proceedings. His research areas include circulating fluidized bed boilers, combined cycle power generation, second law
analysis of thermal systems, and waste heat recovery. He was the recipient of the
President of India medal (1995) from the Institution of Engineers (India). He is a
Fellow of the National Academy of Engineering (FNAE) and a Fellow of the Institution of Engineers (India). He is a Life Member of the Indian Society for Technical
Education, Indian Society for Heat and Mass Transfer, and the Combustion Institute,
USA (Indian section). He was formerly a Member of the New York Academy of
Sciences, USA.
Other books by the same author (all published by Tata McGraw-Hill Publishing
Co. Ltd):
978-0-07-047338-6
Basic and Applied Thermodynamics
978-0-07-059114-1
Engineering Thermodynamics, 3/e
978-0-07-060653-1
Heat and Mass Transfer, 2/e
Fourth Edition
P K Nag
Former Professor
Department of Mechanical Engineering
Indian Institute of Technology
Kharagpur
Tata McGraw-Hill Publishing Company Limited
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Dedicated to the memory of my parents
Late Ananta Lal Nag
Late Jyotsna Rani Nag
Contents
Preface to the Fourth Edition
Preface to the First Edition
Acknowledgements
Nomenclature
Quotes
Before you Begin
xv
xvii
xviii
xix
xxii
xxiii
1. Introduction
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
1.14
1.15
1
Simple Steam Power Plant 2
Internal Combustion (I.C.) Engines 3
Domestic Refrigerator 5
Room Air Conditioner 6
Fuel Cells 7
Macroscopic Versus Microscopic Viewpoint 8
Thermodynamic System and Control Volume 9
Thermodynamic Properties, Processes and Cycles 10
Homogeneous and Heterogeneous Systems 11
Thermodynamic Equilibrium 11
Quasi-Static Process 12
Pure Substance 13
Concept of Continuum 14
Thermostatics 15
Units and Dimensions 15
Solved Examples 20
Summary 22
Review Questions 23
Problems 24
2. Temperature
2.1
2.2
Zeroth Law of Thermodynamics 26
Measurement of Temperature—The Reference Points
26
26
viii
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
Contents
Comparison of Thermometers 28
Ideal Gas 29
Gas Thermometers 29
Ideal Gas Temperature 30
Celsius Temperature Scale 33
Electrical Resistance Thermometer 33
Thermocouple 33
International Practical Temperature Scale
34
Solved Examples 35
Summary 38
Review Questions 38
Problems 39
3. Work and Heat Transfer
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
41
Work Transfer 41
pdV-Work or Displacement Work 42
Indicator Diagram 46
Other Types of Work Transfer 48
Free Expansion with Zero Work Transfer 52
Net Work Done By a System 53
Heat Transfer 53
Heat Transfer—A Path Function 54
Specific Heat and Latent Heat 56
Points to Remember Regarding Heat Transfer and
Work Transfer 56
Solved Examples 57
Summary 64
Review Questions 64
Problems 65
4. First Law of Thermodynamics
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
First Law for a Closed System Undergoing a Cycle 69
First Law for a Closed System Undergoing a Change of State 71
Energy—A Property of the System 72
Different Forms of Stored Energy 72
Specific Heat at Constant Volume 75
Enthalpy 76
Specific Heat at Constant Pressure 76
Energy of An Isolated System 77
Perpetual Motion Machine of the First Kind—PMM1 77
Solved Examples 78
Summary 83
Review Questions 83
Problems 84
69
Contents
ix
5. First Law Applied to Flow Processes
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
87
Control Volume 87
Steady Flow Process 88
Mass Balance and Energy Balance in a Simple
Steady Flow Process 88
Some Examples of Steady Flow Processes 92
Comparison of S.F.E.E. With Euler and Bernoulli Equations 95
Variable Flow Processes 96
Example of a Variable Flow Problem 98
Discharging and Charging a Tank 99
Solved Examples 101
Summary 111
Review Questions 112
Problems 112
6. Second Law of Thermodynamics
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
6.12
6.13
6.14
6.15
6.16
6.17
6.18
117
Qualitative Difference between Heat and Work 117
Cyclic Heat Engine 118
Energy Reservoirs 120
Kelvin-Planck Statement of Second Law 121
Clausius’ Statement of the Second Law 121
Refrigerator and Heat Pump 122
Equivalence of Kelvin-Planck and Clausius Statements 124
Reversibility and Irreversibility 125
Causes of Irreversibility 126
Conditions for Reversibility 130
Carnot Cycle 131
Reversed Heat Engine 133
Carnot’s Theorem 134
Corollary of Carnot’s Theorem 135
Absolute Thermodynamic Temperature Scale 135
Efficiency of the Reversible Heat Engine 139
Equality of Ideal Gas Temperature and Kelvin Temperature 139
Types of Irreversibility 141
Solved Examples 141
Summary 150
Review Questions 151
Problems 153
7. Entropy
7.1
7.2
7.3
157
Introduction 157
Two Reversible Adiabatic Paths Cannot Intersect Each Other 157
Clausius’ Theorem 158
x
Contents
7.4
7.5
7.6
7.7
7.8
7.9
7.10
7.11
7.12
7.13
7.14
7.15
The Property of Entropy 159
Temperature-Entropy Plot 161
The Inequality of Clausius 162
Entropy Change in an Irreversible Process 164
Entropy Principle 166
Applications of Entropy Principle 168
Entropy Transfer with Heat Flow 174
Entropy Generation in a Closed System 175
Entropy Generation in an Open System 178
First and Second Laws Combined 180
Reversible Adiabatic Work in a Steady Flow System 181
Entropy and Direction: The Second Law—a Directional
Law of Nature 183
Entropy and Disorder 183
Absolute entropy 184
Postulatory Thermodynamics 184
7.16
7.17
7.18
Solved Examples 185
Summary 198
Review Questions 199
Problems 200
8. Available Energy, Availability and Irreversibility
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
8.10
8.11
8.12
8.13
209
Available energy 209
Available Energy Referred to a Cycle 209
Quality of Energy 213
Maximum Work in a Reversible Process 215
Reversible Work by an Open System Exchanging
Heat only with the Surroundings 217
Useful Work 221
Dead State 223
Availability 223
Availability in Chemical Reactions 225
Irreversibility and Gouy-Stodola Theorem 227
Availability or Exergy Balance 231
Second Law Efficiency 234
Comments on Exergy 240
Solved Examples 243
Summary 264
Review Questions 265
Problems 267
9. Properties of Pure Substances
9.1
9.2
p-v Diagram for a Pure Substance 273
p-T Diagram for a Pure Substance 278
273
xi
Contents
9.3
9.4
9.5
9.6
9.7
9.8
9.9
p-v-T Surface 279
T-s Diagram for a Pure Substance 280
h-s Diagram or Mollier Diagram for a Pure Substance
Quality or Dryness Fraction 285
Steam Tables 286
Charts of Thermodynamic Properties 288
Measurement of Steam Quality 289
282
Solved Examples 295
Review Questions 314
Problems 314
10. Properties of Gases and Gas Mixtures
Avogadro’s Law 320
Equation of State of a Gas 320
Ideal Gas 323
Equations of State 335
Virial Expansions 336
Law of Corresponding States 337
Other Equations of State 344
Properties of Mixtures of Gases—Dalton’s
Law of Partial Pressures 346
10.9 Internal Energy, Enthalpy and Specific Heats of
Gas Mixtures 349
10.10 Entropy of Gas Mixtures 349
10.11 Gibbs Function of a Mixture of Inert Ideal Gases
320
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8
351
Solved Examples 352
Review Questions 373
Problems 375
11. Thermodynamic Relations, Equilibrium
and Stability
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
Some Mathematical Theorems 384
Maxwell’s Equations 386
TdS Equations 386
Difference in Heat Capacities 387
Ratio of Heat Capacities 389
Energy Equation 390
Joule-Kelvin Effect 393
Clausius-Clapeyron Equation 396
Evaluation of Thermodynamic Properties from an
Equation of State 400
11.10 General Thermodynamic Considerations on an
Equation of State 402
384
xii
11.11
11.12
11.13
11.14
11.15
11.16
Contents
Mixtures of Variable Composition 404
Conditions of Equilibrium of a Heterogeneous System 407
Gibbs Phase Rule 409
Types of Equilibrium 410
Local Equilibrium Conditions 413
Conditions of Stability 414
Solved Examples 415
Review Questions 428
Problems 430
12. Vapour Power Cycles
438
12.1 Simple Steam Power Cycle 438
12.2 Rankine Cycle 440
12.3 Actual Vapour Cycle Processes 444
12.4 Comparison of Rankine and Carnot Cycles 446
12.5 Mean Temperature of Heat Addition 447
12.6 Reheat Cycle 449
12.7 Ideal Regenerative Cycle 451
12.8 Regenerative Cycle 454
12.9 Reheat-Regenerative Cycle 457
12.10 Feedwater Heaters 459
12.11 Exergy Analysis of Vapour Power Cycles 463
12.12 Characteristics of an Ideal Working Fluid in Vapour Power
Cycles 464
12.13 Binary Vapour Cycles 466
12.14 Thermodynamics of Coupled Cycles 468
12.15 Process Heat and By-Product Power 470
12.16 Efficiencies in Steam Power Plant 472
Solved Examples 475
Review Questions 495
Problems 496
13. Gas Power Cycles
13.1 Carnot Cycle (1824) 504
13.2 Stirling Cycle (1827) 505
13.3 Ericsson Cycle (1850) 506
13.4 An Overview of Reciprocating Engines 507
13.5 Air Standard Cycles 508
13.6 Otto Cycle (1876) 508
13.7 Diesel Cycle (1892) 515
13.8 Limited Pressure Cycle, Mixed Cycle or Dual Cycle
13.9 Comparison of Otto, Diesel, and Dual Cycles 519
13.10 Lenoir Cycle 520
13.11 Atkinson Cycle 521
504
517
xiii
Contents
13.12 Brayton Cycle 522
13.13 Aircraft Propulsion 536
13.14 Brayton-Rankine Combined Cycle
540
Solved Examples 543
Review Questions 558
Problems 559
14. Refrigeration Cycles
14.1
14.2
14.3
14.4
14.5
14.6
14.7
14.8
566
Refrigeration by Non-Cyclic Processes 566
Reversed Heat Engine Cycle 567
Vapour Compression Refrigeration Cycle 568
Absorption Refrigeration Cycle 578
Heat Pump System 582
Gas Cycle Refrigeration 583
Liquefaction of gases 584
Production of Solid Ice 586
Solved Examples 587
Review Questions 598
Problems 599
15. Psychrometrics
15.1
15.2
15.3
Properties of Atmospheric Air
Psychrometric Chart 608
Psychrometric Process 610
604
604
Solved Examples 618
Review Questions 628
Problems 629
16. Reactive Systems
16.1 Degree of Reaction 632
16.2 Reaction Equilibrium 635
16.3 Law of Mass Action 635
16.4 Heat of Reaction 636
16.5 Temperature Dependence of the Heat of Reaction 637
16.6 Temperature Dependence of the Equilibrium Constant 638
16.7 Thermal Ionization of a Monatomic Gas 638
16.8 Gibbs Function Change 639
16.9 Fugacity and Activity 642
16.10 Displacement of Equilibrium Due to a Change in
Temperature or Pressure 643
16.11 Heat Capacity of Reacting Gases in Equilibrium 644
16.12 Combustion 645
16.13 Enthalpy of Formation 646
632
xiv
Contents
16.14 First Law for Reactive Systems 647
16.15 Adiabatic Flame Temperature 648
16.16 Enthalpy and Internal Energy of Combustion:
Heating Value 649
16.17 Absolute entropy and the Third Law of Thermodynamics
16.18 Second Law Analysis of Reactive Systems 652
16.19 Chemical Exergy 652
16.20 Second Law Efficiency of a Reactive System 655
16.21 Fuel Cells 655
651
Solved Examples 657
Review Questions 670
Problems 671
17. Compressible Fluid Flow
17.1
17.2
17.3
17.4
17.5
17.6
676
Velocity of Pressure Pulse in a Fluid 676
Stagnation Properties 678
One Dimensional Steady Isentropic Flow 680
Critical Properties—Choking in Isentropic Flow
Normal Shocks 687
Adiabatic Flow with Friction and Diabatic Flow
without Friction 693
681
Solved Examples 694
Review Questions 702
Problems 703
18. Gas Compressors
18.1
18.2
18.3
18.4
18.5
18.6
18.7
Compression Processes 706
Work of Compression 706
Single-Stage Reciprocating Air Compressor
Volumetric Efficiency 711
Multi-Stage Compression 713
Air Motors 719
Rotary Compressor 719
706
709
Solved Examples 723
Review Questions 733
Problems 734
Appendix A
Appendix B
Appendix C– Multiple Choice Questions
Bibliography
Index
737
787
788
796
797
Preface to the Fourth Edition
To bring out a new edition gives an opportunity to the author to sift through the text
and to rearrange, add and remove any material. The book was copiously reviewed
again by eminent teachers on the subject. Based on their suggestions, some modifications have been carried out. A small introductory note on the contributions of
prominent founders of the subject is provided. Some engineering applications of
thermodynamics is furnished to create interest on the subject. An alternative method
is also given to prove the inequality of Clausius, which is not much different from
what was given in the original text. The summary of chapters has been given for
each up to Chapter 8, which form the basic principles of the subject. A number of
multiple choice questions with answers are furnished at the end to help the students
in competitive examinations. Some more numerical problems are provided, either
solved or given in exercises. The chapter on ‘Elements of Heat Transfer’ is eliminated since it is now offered in many disciplines as a separate course. In spite of all
these alterations, the book retains the broad generality and simplicity of the subject.
Thermodynamics, being highly conceptual rather than mathematical, is often
characterized as a difficult subject. The book attempts to urge the students to adopt
a fundamental approach as outlined in the book, work to understand the concepts
and develop the ability to apply the basic principles in a systematic way. This
approach will make the students feel at ease with the subject and it will equip them
with a tremendously useful set of tools for engineering analysis.
The accompanying website of the book can be accessed at http://www.mhhe.
com/nag/et4e and contains the following material:
n For Instructors
· Solution Manual
· PowerPoint lecture slides
n For Students
· web links for additional reading
I am greatly thankful to Ms Surabhi Shukla for her arduous task of syllabi
research and competitive analysis which helped me a great deal in preparing this
revision. I would also profusely thank Ms Sohini Mukherjee of Tata McGraw-Hill
for her deft editorial services. Mr P L Pandita too deserves a special thanks for his
efficient handling of the production process. I am also thankful to all the following
reviewers of the book.
xvi
Preface to the Fourth Edition
S N Garg
Dept. of Mechanical Engineering, Engineering
College and Research, Jodhpur
C P Jain
Dept. of Mechanical Engineering, University
College of Engineering, Kota
Ajay Trehan
Dept. of Mechanical Engineering, National
Institute of Technology, Jalandhar
M K Paswan
Dept. of Mechanical Engineering, National
Institute of Technology, Jamshedpur
T K Jana
Dept. of Mechanical Engineering, Haldia Institute
of Technology, Haldia
P K Bardhan
Dept. of Mechanics, JIS College of Engineering,
Nadia (W B)
D A Warke
Dept. of Mechanical Engineering, J T Mahajan
College of Engineering, Jalgaon
P Chidhambaranathan
Dept. of Mechanical Engineering, Dr Sivanti
Aditanar College of Engineering, Tutucorin
District, Tamil Nadu
M S Senthil Saravanan
Dept. of Mechanical Engineering, Indian
Engineering College, Tamil Nadu
Dr J Pattabiraman
M N M Jain College, Thorapakkam, Chennai
A V S Gupta
Dept. of Mechanical Engineering, College of
Engineering, JNTU, Hyderabad
Y. S. Varadarajan
Dept. of Industrial and Production Engineering,
The National Institute of Engineering, Mysore.
I am grateful to Ms Vibha Mahajan and Ms Shukti Mukherjee of Tata McGrawHill for overall supervision, organization and production of this revised edition.
The author hopes that the book would fit all curriculums of basic engineering
thermodynamics and would remain as popular as it has ever been. Any constructive
suggestion for further improvement of the book would be welcome and gratefully
acknowledged.
P K NAG
Preface to the First Edition
The purpose of this book is to provide a mature approach to the basic principles of
classical thermodynamics which is one of the few core subjects for the undergraduate students of almost all branches of engineering. The system-surroundings interactions involving work and heat transfer with associated property changes, and the
system-control volume approaches of the first law have been emphasized. The second law has been elaborated upon in considerable detail. Except for some physical
explanations, a statistical or microscopic analysis of the subject has not been made.
The first eight chapters of the book are devoted to a thorough treatment of the
basic principles and concepts of classical thermodynamics. The second law and
entropy have been introduced using the concept of heat engine. Chapters 9 and 10
present the properties of substances. Chapter 11 gives the general thermodynamic
relationships among properties. A detailed analysis of power and refrigeration
cycles is given in Chapters 12 to 14. Chapter 15 deals with air-water vapour mixtures, and reactive systems are analyzed in Chapter 16. To increase the utility of the
book, Chapters 17 and 18 dealing with compressible fluid flow and heat transfer
respectively, have been added.
Many illustrative examples are solved and many problems are provided in each
chapter to aid comprehension and to stimulate the interest of the students.
Throughout the text SI units have been used. Tables and charts given in the
Appendix are also in SI units.
This book is based mainly on the lecture notes prepared for classes on the subject at IIT Kharagpur. I am grateful to the authors of the books that I used in preparing the notes, a list of which is given in the bibliography. I am thankful to my
colleagues in the mechanical engineering department of IIT Kharagpur, for many
stimulating discussions, and for encouragement. I am indebted to all those who
have helped in the preparation of the book.
I would very much appreciate criticisms, suggestions for improvement, and
detection of errors from my readers, which will be gratefully acknowledged.
P K NAG
Acknowledgements
The author gratefully remembers the contributions made by Late Prof. Bindu
Madhav Belgaumkar, the Head of the Department of Mechanical Engineering
(1954–1970), at Indian Institute of Technology (IIT), Kharagpur, in building up his
teaching career during his earlier formative years, by providing the right academic
environment in the campus. He also humbly acknowledges the contribution of his
teacher, Prof. R G Mokadam for maintaining high standard of teaching Thermodynamics in the decades of 1950s and 1960s at IIT, Kharagpur, the first IIT which
played the pioneering role in technical education in the country.
The author is indebted to his erstwhile colleagues at IIT for many positive interactions, some of whom were his teachers during his M.Tech course, viz., Prof. C N
Lakshminarayan, Late Prof. H R Narayan, Prof. Jayanand H. Hiranandani, Prof. H
V Rao, Prof. B G Ghosh, Prof. A K Roy, Late Prof. P K G Pannikar, Prof. Darshan
Lal, Prof. N S Murthy and Prof. S K Som.
P K NAG
Nomenclature
The symbols listed here are not exhaustive, but are the main ones used in the text.
Other symbols have been defined at appropriate places in the book.
Symbols
a
af
A
A
b
B
c
cp
cv
Cp
CV
COP
d, D
e
E
f
F
Activity: Specific
availability or exergy, kJ/kg
Exergy of a stream per unit
mass, kJ/kg
Availability or exergy kJ
area, m 2
Specific Keenan function,
kJ/kg
Keenan function, kJ
Velocity of sound, m/s
Specific heat at constant
pressure, kJ/kgK
Specific heat at constant
volume, kJ/kgK
Heat capacity at constant
pressure, kJ/K
Heat capacity at constant
volume, kJ/K
Coefficient of performance
Diameter
Specific energy, kJ/kg
Total energy, kJ
Fugacity; Specific
Helmholtz function, u – Ts,
kJ/kg
Force N
F
g
g
G
h
hf
H
HP
HR
h RP
HHV
i
I
I
K
K
k
kT
Helmhotz function, kJ
Gravitational acceleration,
m/s2
Specific Gibbs function,
h – Ts, kJ/kg
Total Gibbs function, kJ
Specific enthalpy, kJ/kg
Enthalpy of formation,
kJ/kg mol
Total enthalpy, U + pV, kJ
Enthalpy of products, kJ
Enthalpy of reactants, kJ
Enthalpy of combination,
kJ/kg fuel
Higher heating value, kJ/kg
fuel
Specific irreversibility, kJ/kg
Total irreversibility, kJ
Electric current, A
Boltzmann constant,
J/molecule–K.
Indicator spring constant,
N/cm3
Thermal conductivity,
W/mK
Isothermal compressibility,
K–1
xx
ks
K
KE
l
L
LHV
Nomenclature
Adiabatic compressibility,
m2/N
Equilibrium constant
Total kinetic energy, kJ
Latent heat, kJ/kg
Length, m
Lower heating value, kJ/kg
fuel
Mass, kg
m
m&
Mass flow rate, kg/s
M
Mach number
M
Molar mass, kg/kg mol
M EP, mep Mean effective pressure,
N/m2
mf
Mass fraction
n
Polytropic index, number of
moles
N
Number of molecules
p
Pressure, N/m2 (Pa)
p
Partial pressure, Pa
pc, pcr
Critical pressure, kPa
po
Atmospheric pressure, kPa
pr
Reduced pressure
pv
Vapour pressure, kPa
PE
Total Potential energy, kJ
Q
Total heat transfer, kJ
Q
Heat transfer rate kJ/s
QH , Q1
Heat transfer from high
temperature body, kJ
QL , Q2
Heat transfer to low
temperature body, kJ
Qrev
Heat transferred and
reversibly, kJ
rc
Cut-off ratio
re
Expansion ratio
gk
Compression ratio
gp
Constant volume pressure
ratio
R
Characteristic gas constant,
kJ/kg – K
R, R u
Universal gas constant, kJ/kg
mol – K
s
S
Siso
t
tf
T
Tas
Tcr, Tc
Tdb
Tdp
Twb
TH, T1
TL, T2
T0
Tr
u
U
v
vc, vcr
vr
V
W
W&
Wmax
Wrev
Wu
x
x
z
Z
Specific entropy, kJ/kg – K
Total entropy, kJ/K –
Siso, Suniv Entropy of the
isolated system or universe,
kJ/K
Time, and; temperature, ºC
Final temperature, ºC
Temperature K
Adiabatic saturating
temperature, ºC or K.
Critical temperature, K
Dry bulb temperature, ºC or
K
Dew point temperature ,ºC
Wet bulb temperature, ºC
Temperature of hot reservoir
or source, K
Temperature of cold
reservoir or sink, K
Temperature of the
surroundings, K
Reduced temperature
Specific internal energy,
kJ/kg
Total internal energy, kJ
Specific volume, m3/kg
Critical specific volume, m3/kg
Reduced volume
Total volume, m3
Total work, kJ
Power, kW
Maximum work, kJ
Reversible work, kJ
Useful work, kJ
Quality, dryness fraction
Mass fraction, Mole fraction
Elevation, m
Compressibility factor
Greek Letters
b
D
Volume expansivity, K–1
Finite change in quantity
xxi
Nomenclature
e
h th
hI
h II
m
mJ
n
r
s
s
t
t
f
F
y
w
Effectiveness
Thermal efficiency
First law efficiency
Second law efficiency
Chemical potential, kJ/kg
Joule-Thomson coefficient,
K/kPa
Stoichiometric coefficient
Density, kg/m3
Stefan-Boltzmann
coefficient
Surface tension, N/m
Torque, Nm
Time, s
Relative humidity
Availability or energy of a
closed system, kJ/kg
Availability or exergy of a
steady flow system, kJ/kg
Specific humidity or
humidity ratio, kg water
vap/kg dry air
Subscripts
a
abs
act
atm
av
cr
CV
Air
Absolute
Actual
Atmospheric
Average
Critical point
Control volume
e
f
fg
m
r
rev
s
sat
surr
sys
v
0
Exit conditions
Saturated liquid
Difference in property
between saturated liquid and
saturated vapour
Saturated vapour
Generation
High temperature (as in TH
and QH)
Inlet conditions
i-th component
Low temperature (as in TL
and QL)
Mixture mean
Reduced
Reversible
Isentropic
Saturated
Surroundings
system
Water vapour
Dead state
1
2
Initial or inlet state
Final or exit state
g
gen
H
i
i
L
Superscripts
. (over dot) Quantity per unit time
(over bar) Quantity per unit mol
º (circle) Standard reference state
¢ (one prime) Solid phase
¢¢ (two prime) Liquid phase
¢¢¢ (three prime) Vapour phase
–
Quotes
A theory is more impressive if the greater is the simplicity of its premises, the more
different are the things it relates, and the more extended its range of applicability.
Therefore, the deep impression which classical thermodynamics made on me. It is
the only physical theory of universal content which I am convinced, that within the
framework of applicability of its basic concepts will never be overthrown.
— Albert Einstein
There is nothing more practical than a good theory.
— Ludwig Boltzmann
Thermodynamics is a funny subject. The first time you go through it, you don’t
understand it at all. The second time you go through it, you think you understand it,
except one or two points. The third time you go through it, you don’t understand it
but by that time you are so used to the subject, it doesn’t bother you any more.
— Arnold Sommerfeld
It is a remarkable illustration of the ranging power of the human intellect that a
principle first detected in connection with the clumsy puffing of the early steam
engines should be found to apply to the whole world, and possibly, even to the
whole cosmic universe.
— A R Ubbelohde
The energy of the world is constant. The entropy of the world tends towards a
maximum.
— Rudolf Clausius
The increase of disorder or entropy with time is one example of what is called an
arrow of time something that gives a direction to time and distinguishes the past
from the future. There are at least three different directions of time. First, there is
the thermodynamic arrow of time—the direction of time in which disorder or
entropy increases. Second, there is the psychological arrow of time. This is the
direction in which we feel time passes—the direction of time in which we remember the past, but not the future. Third, there is the cosmological arrow of time. This
is the direction of time in which the universe is expanding rather than contracting.
Before You Begin
— Stephen W. Hawking
Thermodynamics, a classical or macroscopic science, studies various energy interactions, notably heat and work transfer, with matter, which bring about changes in
the macroscopic properties of a substance that are perceptible and measurable. It is
a phenomenological science based on certain laws of nature which are always
obeyed and never seen to be violated. The first law introduces energy as a property
and emphasizes the conservation principle. Energy is neither created nor destroyed
with reference to an isolated system or universe. It only gets transformed from one
form to another. The first law does not discriminate on the kind of energy, the unit
of all forms being kJ. It is the second law which states that all kinds of energy are
not of the same ‘quality’. Electricity or work is a superior form of energy, while
heat is an inferior form. Exergy is a measure of the quality index of energy. Professor Joseph H Keenan of MIT first introduced the term ‘availability’ to indicate it in
1941, which became popular in the USA and the UK. The German engineer Rant
first used the term ‘exergy’ in 1956, and it received global acceptance, since it is
closer to energy in spelling and pronunciation.
Exergy or availability (A) of a given system is defined as the maximum useful
work that is obtainable in a process in which the system comes to equilibrium with its
surroundings, i.e., the ‘dead state’ (p 0, t 0). The exergy is thus a composite property
depending on the state of both system and the surroundings. Any change in the state of
the system from the dead state is a measure of the exergy or work that can be extracted
from it. The farther the initial state of the system from the dead state in terms of p, t
either above or below it, the higher will be the exergy of the system (Fig. 1). All
spontaneous processes terminate at the dead state, where the exergy is zero.
1
p0
1
De
p
O
O
T
p0
Dead state
T0
Dead
state
T0
1¢
1¢
v
s
Fig. 1 Exergy of a system decreases as its state approaches the dead state at ‘p0 , T0
xxiv
Before You Begin. . .
The maximum work or exergy of a system per unit mass in a steady flow process
is given by
V2
+ gz) – (h0 – T0 s0 + gz)
2
If subscripts 1 and 2 denotes the states of the system entering and leaving a CV,
the decrease in exergy is
a = (h – T0 s +
Wmax = a1 – a2 = (h1 – T0 s1) – (h2 – T0 s2 )
= b1 – b2
where b is the specific Keenan function. Here KE, PE terms are neglected.
The exergy of a closed system or the availability in a nonflow process is
A = Wmax = (E – E0) + p0(V – V0) – T0(S – S0)
MV 2
+ mgz) – (U0 + mgz 0) + p0(V – V0) – T0(S – S0)
2
If KE and PE terms are neglected, and for unit mass, it is
a = u – u0 + p0(v – v0) – T0(s – s0).
= (U +
Irreversibility and Gouy–Stodola Theorem
The actual work done by a system is always less than the idealized reversible work,
and the difference between the two is called the irreversibility.
I = Wmax – W
This is also called ‘degradation’ or ‘dissipation’.
In a nonflow process between two equilibrium states, when the system exchanges
heat only with the environment
I = [(U1 – U2) – T0(S1 – S2)] – [(U1 – U2) + Q]
= T0(S 2 – S1) – Q = T0(DS)sys + T0(DS)surr
= T0(DS)univ = T0Sgen
Since Sgen is always positive, I > 0.
Similarly, for a steady flow process
I = Wmax – W
= [(B1 +
mV12
mV22
+ mgz 1) – (B2 +
+ mgz 2)]
2
2
mV12
mV22
+ mgz 1) – (H2 +
+ mgz 2 )]
2
2
= T0(S 2 – S1) – Q = T0(DS)sys + T0(DS)surr
– [(H1 +
= T0(DS)univ = T0Sgen.
The quantity T0 Sgen represents a decrease in exergy or exergy destruction.
The Gouy-Stodola theorem states that the rate of loss of exergy in a process is
º
proportional to the rate of entropy generation, S gen , or
xxv
Before You Begin. . .
º
º
I = T0 S gen
A thermodynamically efficient process would involve minimum exergy loss with
minimum rate of entropy generation.
When a closed system is allowed to undergo a spontaneous change from the given
state to the dead state, the exergy is completely destroyed without producing any
useful work.
Exergy Balance for a Closed System
For a closed system, exergy transfer occurs through heat and work interactions (Fig. 2).
2
1st law
E2 – E1 = ò dQ – W1 – 2
(1)
Q
Ts
1
2
2nd law
dQ
= Sgen
T
1
S2 – S1 – ò
1Æ2
2
or
dQ
= T0 S gen
T
1
T0(S2 – S 1) – T0 ò
(2)
Boundary
W1–2
Fig. 2
From equations (1) and (2),
2
2
dQ
– T0 Sgen
T
1
E2 – E1 – T0(S 2 – S1) = ò dQ – W1 – 2 = T0 ò
1
2
æ
T ö
ò çè1 - T0 ÷ø dQ – W1 –2 – T0 Sgen
1
2
æ T ö
A2 – A1 = ò ç1 - 0 ÷ dQ – [W1 – 2 – p0(V2 – V1) – T0 Sgen
è Tø
(3)
1
For an isolated system), the exergy balance gives
A2 – A1 = – T0 Sgen = – I
(4)
Since I > 0, the only processes allowed by the second law are those for which
exergy of the isolated system decreases.
The exergy of an isolated system can never increase. It is the counterpart of the
entropy principle which states that entropy of an isolated system can never decrease.
Similarly, for a steady flow process,
I&
W&
æ T0 ö Q&
çè1 - T ÷ø m + a f1 - m& - a f 2 = m&
s
Exergy in
Exergy out
Exergy loss
Second-Law Efficiency
By first law,
Energy input – Energy output = Energy loss
First law efficiency, hI =
Energy output
Energy input
xxvi
Before You Begin. . .
By second law,
Exergy input – Exergy output = Exergy loss or Irreversibility
Second law efficiency, h II =
Exergy output
Exergy input
It can also be defined as
hII =
A
W
= min (for a cycle)
Wmax
A
æ T ö
Since Wmax = A = Q1 ç1 - 0 ÷
è T1 ø
hII =
W
æ T ö
Q1 ç1 - 0 ÷
è T1 ø
=
hI
h carnot
æ T ö
If work is involved, Amin = W(desired), and if heat is involved, Amin = Q ç1 - 0 ÷ .
è T1 ø
Introduction
1
4/25/08, 9:45 AM
* P.K. Nag, “Basic and Applied Thermodynamics,” Tata McGraw-Hill, New Delhi, 2002,
page 13.
Thermodynamics is the science of energy transfer and its effect on the physical
properties of substances. It is based upon observations of common experience
which have been formulated into thermodynamic laws. These laws govern the
principles of energy conversion. The applications of the thermodynamic laws and
principles are found in all fields of energy technology, notably in steam and nuclear
power plants, internal combustion engines, gas turbines, air conditioning,
refrigeration, gas dynamics, jet propulsion, compressors, chemical process plants,
and direct energy conversion devices.
One may be curious to know about the founders of engineering thermodynamics.
A list of individuals whose lives were vital to the founding of the subject is given
below with some information in regard to their contributions.
One thermodynamicist, Antoine Laurent Lavoisier (1743–1794), was the father
of modern chemistry and was particularly interested in the compounding of gunpowder. It was Lavoisier’s unfortunate fate that he lived during the French Revolution (1789). Before the revolution he was a recognised scientist and contributed to
the analysis of the combustion process. He led in developing the concept of elements; he proposed the name of ‘oxygen’ (discovered by Joseph Priestley) and
described its function in the combustion process. He was born into a prosperous
middle-class family and pursued an education in the sciences. He was active in
politics before the revolution, seeking greater social reforms. He pursued various
scientific topics including the development of the metric system. However, his role
as a tax collector did not endear him to the public and he was beheaded by the
guillotine.
Another foremost contributor was Nicolas Leonard Sadi Carnot (1796–1832),
whose observations led to the naming of a theoretical thermodynamic cycle after
him. He lived his youth in Napoleon’s France, and his father, one of Napoleon’s
military organizers, was able to provide Carnot, with a sound technical education
(see * below). He left the army at the age of 24 and moved to Pairs to do work in
1
2
Engineering Thermodynamics
thermodynamics and wrote the famous monograph ‘Reflections on the Motive
Power of Fire’ (1824) and established the basic principles governing the conversion of heat to work that led to the second law of thermodynamics.
James Prescott Joule (1818–1889), a British scientist of the mid-nineteenth century, performed careful and throughtful experiments to establish heat as a form of
energy refuting the calorific theory of heat. It led to the founding of the first law of
thermodynamics with experimental proof of heat as a form of energy.
William Thomson (1824–1907), who later became Lord Kelvin, a renowned British physicist, developed the thermodynamic temperature scale. The absolute temperature scale was named after him. From his experiments he deduced the doctrine
of available energy, and along with Joule expounded the Joule–Kelvin effect.
A German professor, Rudolph J. Clausius (1822-1888), another famous contributor in thermodynamics, used some of Carnot’s reflections and spent the mid1860s developing ideas about matter and publishing papers dealing with an
abstract quantity called ‘entropy’.
The last thermodynamics pioneer we will discuss is J. Willard Gibbs (1839–
1903), a comparatively unknown American, who contributed tremendously in various areas of thermodynamics, statistical mechanics, chemistry and mathematics.
He studied and taught at Yale University, and being disreputed for his odd ideas
worked for nine years without pay. Eventually his worth was recognised in academic circles, and he was paid the salaries later.
The strength of thermodynamics lies in its ability to analyze and design a wide
range of systems using only a few tenets—the four laws of Thermodynamics. Some
common applications of work-producing and work-consuming plants are being
described below.
1.1
SIMPLE STEAM POWER PLANT
A schematic diagram of a simple steam power plant is shown in Fig. 1.1. High
pressure, high-temperature superheated steam leaving the steam generator or boiler
does work in the turbine which drives the electric generator to produce electricity.
The low-pressure steam from the turbine enters the condenser, where heat is transferred from the steam to the cooling water circulating through the tubes. Steam is
condensed to water which is then pumped back to the boiler.
A fossil fuel (coal, fuel oil or natural gas) is used to release heat by combustion,
and the products of combustion or flue gases heat water to generate steam and then
superheat it. The exhaust gases, before leaving through stack or chimney, heat the
feedwater in a heat exchanger called economizer before the feedwater enters the
steam drum and then heats the air for combustion in another heat exchanger called
air preheater. The exhaust gases leaving the stack have a considerable amount of
greenhouse gases (CO2) and cause thermal pollution and global warming. A large
power plant has many other pieces of equipment, some of which will be considered later.
3
Introduction
In a nuclear power plant, the nuclear reactor replaces the furnace of a conventional power plant. The turbine, condenser and pump are the three other basic components. Safety considerations and the disposal of the radioactive wastes are important in the design and operation of a nuclear plant.
Stack
gases out Air in
Air
preheater
High-pressure
superheated steam
Economizer
Hot water
Turbine
Generator
Warm air
Low-pressure steam
Superheater
Condenser
Fuel
Steam generator
Cooling water out
Pump
High-pressure
low-temperature
water to boiler
Cooling water from
river or lake or
cooling tower
Fig. 1.1
1.2
Pump
Schematic diagram of a simple steam power plant.
INTERNAL COMBUSTION (I.C.) ENGINES
These are largely used in transport vehicles like cars, buses, trucks, motor cycles,
and so on. Here, a fuel is burned, and the energy from the burning fuel is transferred to the pistons, which through gears, turn the wheels, thus moving the automobile (Fig. 1.2).
In petrol engines, the fuel-air mixture after being compressed is ignited by an
electric spark, hence the name spark ignition (S.I.) engine, and the products of
combustion do work on the piston, and through crank-and-connecting-rod mechanism, power is transferred to the crank shaft. In diesel engines, only air is inducted
by suction into the cylinder and compressed to a high pressure. The fuel is injected
in fine atomized form into the hot compressed air. The mixture gets self-ignited and
the combustion products do work in the pistons. These are called compression ignition (C.I.) engines.
Thermodynamic analysis seeks to determine how much work we may expect
from an engine and, through experiments, how efficiently the engine is performing. This is very important if the pollution from exhausts is to be minimized.
4
Engineering Thermodynamics
Throttle
Spark� plug
Rocker� arm
Exhaust� manifold
Fuel� supply
To� atmosphere
Air� supply
Exhaust� valve
Carburetor
Intake� manifold
Combustion� chamber
Piston
Intake� valve
Cylinder
Pushrod
Connecting� rod
Camshaft
Crankcase
Crankshaft
Oil� pan
(a) Reciprocating mechanism for petrol and diesel engines
Cylinmder
Suction
valve
Connecting
rod
Crank
(Crank� shaft
rotary� motion)
Exhaust
valve
Piston� (Reciprocating� motion)
Piston� rings
(b) An automotive engine
Fig. 1.2 Internal combustion engine
The gas turbine (Fig. 1.3) is another automotive power source, more commonly
found in jet planes. There is an upsurge in the development of gas turbine plants in
both electric power generation and ship propulsion. Air is compressed and energy
added to it by burning fuel in a combustion chamber; this mixture, viz., the
5
Introduction
products of combustion, expands through a turbine, doing
work, which drives the electric
generator or the ship. The
analysis is similar to that of
most power plants, and all
these analyses have a common
purpose, which is to consider
how efficiently the chemical
energy of the fuel is converted
into mechanical energy. The
processes of converting the
energy are different, but the
principle of energy conversion
remains the same.
Air� inlet
Starting
motor
Gas
turbine
Compressor
Exhaust
Fuel
Fig. 1.3 A gas turbine unit
1.3 DOMESTIC REFRIGERATOR
A schematic diagram of the refrigerator is shown in Fig. 1.4. The basic components
are (a) Evaporator, (b) Compressor, (c) Condenser, (d) Expansion device. The
evaporator where the refrigerant (working fluid) evaporates absorbing the latent
heat of vaporization is the coldest part of the freezer cabinet, where it is located. In
modern frost-free refrigerators, the evaporator is located outside the cabinet. A fan
circulates air from the evaporator to the freezer. Just below the freezer, there is
a chiller tray. Further below are compartments with progressives higher temperatures. The bottom-most compartment meant for vegetables is the least cold
one. The cold air being heavier flows down from the freezer to the bottom of the
Refrigerator� cabinet
Air� out
Evaporator
(Freezer)
QL
Condenser
QH
Capillary
tube
Air� in
W
Compressor
Fig. 1.4 Schematic diagram of a domestic refrigerator
6
Engineering Thermodynamics
refrigerator. The warm air being lighter flows upward from the vegetable box to the
freezer, gets cooled and flows down again. Thus, a natural convection current is set
up which maintains a temperature gradient between the top and bottom of the refrigerator. The temperature maintained in the freezer is about – 15°C. The condenser is usually a wire-and-tube type mounted at the back of the refrigerator, having no fan. The refrigerant vapour is condensed with the help of the surrounding air
which rises above by natural convection as it gets heated after absorbing the latent
heat of condensation from the refrigerant. After condensation, the high pressure
liquid refrigerant is reduced to the low pressure of the evaporator by passing through
an expansion device (throttle valve or capillary tube) and the cycle is completed.
Earlier, the refrigerant used was R-12 and because of the ozone layer depletion it
is now isobutene (C4 H10 ), propane (C3 H8 ) or tetrafluoroethane (C2H2F4 or
R-134a).
1.4 ROOM AIR CONDITIONER
The schematic diagram of a typical window type air conditioner is given in
Fig. 1.5. Let us consider a room to be maintained at a constant temperature of
25°C. The air from the room is drawn by a fan and is made to pass over a cooling
coil of the evaporator, the surface of which is maintained at a temperature of 10°C.
Heated� air� at
55°C
Outside� air
45°C
Condenser
Higher� pressure
liquid� at� 60� °C
High� pressure
vapour
Outside� air
at� 45� °C
Partition� wall
Electric� motor
W
Fan� motor
Return� air
at� 25� °C
Low� pressure� low
temperature� liquid
at� 5-10°C
Evapansion
device
Compresser
Low� pressure
vapour� at� 10-20� °C
Evaporator� (cooling� coil)
Return� air
at� 25� °C
Fig. 1.5
Supply� air� to
room� at� 15°C
Schematic diagram of a room air-conditioner
7
Introduction
After passing over the coil, the air is cooled to around 10°C before being supplied
to the room at 25°C by the fan. In the cooling coil, the refrigerant, R–22 or
R–134a enters at, say, 5°C and evaporates absorbing the latent heat of vaporization
from the room air.
The refrigerant vapour from the evaporator is compressed to a high pressure
before entering the condenser where the atmospheric air at about say, 45°C, in
summer is circulated by a fan. After picking up the latent heat of condensation
from the refrigerant, the air is thrown back to the atmosphere, say, at 55°C. The
high-pressure liquid refrigerant from the condenser is reduced to the low evaporator pressure by passing through the expansion device (capillary tube) before
entering the evaporator. The cycle repeats itself.
1.5 FUEL CELLS
In a conventional power plant, fuel and air enter the power plant and products of
combustion leave the unit. There is also a transfer of heat to the cooling water and
work is done in the form of electrical energy. The fuel cell accomplishes this objective converting the chemical energy of fuel directly to electricity. Figure 1.6 shows
a schematic arrangement of a fuel cell of ion-exchange membrane type. In this fuel
cell hydrogen and oxygen react to form water. The flow of electrons in the external
circuit is from anode to cathode. Hydrogen enters at the anode side, and oxygen
Load
4e–
Anode
Catalytic
electrodes
4e–
–
+
Cathode
Ion-exchange
membrane
Gas� chambers
Hydrogen
Oxygen
4e–
4e–
O2
2H2
4H+
4H+
2H2O
H 2O
Fig. 1.6
Schematic arrangement of an ion-exchange membrane type of fuel cell
8
Engineering Thermodynamics
enters at the cathode side. At the surface of the ion-exchange membrane the hydrogen is ionized according to the reaction
2H2 Æ 4H+ + 4e–
The electrons flow through the external circuit and the hydrogen ions flow
through the membrane to the cathode, where the following reaction takes place:
4H+ + 4e– + O 2 Æ 2H2O
There is a potential difference between the anode and the cathode, and thus there
is a flow of electricity.
At present, the fuel used in fuel cells is either hydrogen or a mixture of gaseous
hydrocarbons and hydrogen. The oxidizer is usually oxygen. However, current
development is directed toward the production of fuel cells that use hydrocarbon
fuels and air. The fuel cell is already being used to produce power for space and
other special applications.
Thermodynamics plays a vital role in the analysis, development and design of
all power-producing systems, including I.C. engines and gas turbines as well as
power-absorbing systems like refrigerators and air conditioners. Considerations
such as the increase of efficiency, improved design, optimum operating conditions,
environmental pollution and alternative methods of power generation, involve,
among other factors, the careful application of the fundamentals of thermodynamics.
1.6 MACROSCOPIC VERSUS MICROSCOPIC VIEWPOINT
There are two points of view from which the behaviour of matter can be studied:
the macroscopic and the microscopic. In the macroscopic approach, a certain
quantity of matter is considered, without the events occurring at the molecular
level being taken into account. From the microscopic point of view, matter is
composed of myriads of molecules. If it is a gas, each molecule at a given instant
has a certain position, velocity, and energy, and for each molecule these change
very frequently as a result of collisions. The behaviour of the gas is described by
summing up the behaviour of each molecule. Such a study is made in microscopic
or statistical thermodynamics. Macroscopic thermodynamics is only concerned
with the effects of the action of many molecules, and these effects can be perceived
by human senses. For example, the macroscopic quantity, pressure, is the average
rate of change of momentum due to all the molecular collisions made on a unit
area. The effects of pressure can be felt. The macroscopic point of view is not
concerned with the action of individual molecules, and the force on a given unit
area can be measured by using, e.g. a pressure gauge. These macroscopic
observations are completely independent of the assumptions regarding the nature
of matter. All the results of classical or macroscopic thermodynamics can, however,
be derived from the microscopic and statistical study of matter.
9
Introduction
1.7 THERMODYNAMIC SYSTEM AND CONTROL VOLUME
A thermodynamic system is defined as a quantity of matter or a region in space
upon which attention is concentrated in the analysis of a problem. Everything
external to the system is called the surroundings or the environment. The system is
separated from the surroundings by the system boundary (Fig. 1.7). The boundary
may be either fixed or moving. A system and its surroundings together comprise a
universe.
There are three classes of systems: (a) closed system, (b) open system, and
(c) isolated system. The closed system (Fig. 1.8) is a system of fixed mass. There is
Boundary
Boundary
Energy out
System
System
Surroundings
Energy in
Surroundings
Fig. 1.7
No mass transfer
A Thermodynamic System
Fig. 1.8
A Closed System
no mass transfer across the system boundary. There may be energy transfer into or
out of the system. A certain quantity of fluid in a cylinder bounded by a piston
constitutes a closed system. The open system (Fig. 1.9) is one in which matter
crosses the boundary of the system. There may be energy transfer also. Most of the
engineering devices are generally open systems, e.g. an air compressor in which
air enters at low pressure and leaves at high pressure and there are energy transfers
across the system boundary. The isolated system (Fig. 1.10) is one in which there is
no interaction between the system and the surroundings. It is of fixed mass and
energy, and there is no mass or energy transfer across the system boundary.
Energy in
Boundary
Mass out
System
System
Surroundings
Surroundings
Mass in
Fig. 1.9
Energy out
An Open System
No mass or energy transfer
Fig. 1.10
An Isolated System
If a system is defined as a certain quantity of matter, then the system contains the
same matter and there can be no transfer of mass across its boundary. However, if a
system is defined as a region of space within a prescribed boundary, then matter can
cross the system boundary. While the former is called a closed system, the latter is
an open system.
10
Engineering Thermodynamics
For thermodynamic analysis of an open system, such as an air compressor
(Fig. 1.11), attention is focussed on a certain volume in space surrounding the
compressor, known as the control volume, bounded by a surface called the control
surface. Matter as well as energy crosses the control surface.
Air out
Heat
Work
Air compressor
Motor
Control
surface
Control volume
Air in
Fig. 1.11 Control Volume and Control Surface
A closed system is a system closed to matter flow, though its volume can change
against a flexible boundary. When there is matter flow, then the system is considered
to be a volume of fixed identity, the control volume. There is thus no difference
between an open system and a control volume.
1.8 THERMODYNAMIC PROPERTIES, PROCESSES AND
CYCLES
Every system has certain characteristics by which its physical condition may be
described, e.g. volume, temperature, pressure, etc. Such characteristics are called
properties of the system. These are all macroscopic in nature. When all the
properties of a system have definite values, the system is said to exist at a definite
state. Properties are the coordinates to describe the state of a system. They are the
state variables of the system. Any operation in which one or more of the properties
of a system changes is called a change of state. The succession of states passed
through during a change of state is called the path of the change of state. When the
path is completely specified, the change of state is called a process, e.g. a constant
pressure process. A thermodynamic cycle is defined as a series of state changes
such that the final state is identical with the initial state (Fig. 1.12).
Properties may be of two types. Intensive properties are independent of the mass
in the system, e.g. pressure, temperature, etc. Extensive properties are related to
mass, e.g. volume, energy, etc. If mass is increased, the values of the extensive
properties also increase. Specific extensive properties, i.e. extensive properties per
unit mass, are intensive properties, e.g. specific volume, specific energy, etc.
11
Introduction
a
p
1
b
2
v
a-b A process
1-2-1 A cycle
Fig. 1.12 A Process and a Cycle
1.9 HOMOGENEOUS AND HETEROGENEOUS SYSTEMS
A quantity of matter homogeneous throughout in chemical composition and
physical structure is called a phase. Every substance can exist in any one of the
three phases, viz. solid, liquid and gas. A system consisting of a single phase is
called a homogeneous system, while a system consisting of more than one phase is
known as a heterogeneous system.
1.10
THERMODYNAMIC EQUILIBRIUM
A system is said to exist in a state of thermodynamic equilibrium when no change
in any macroscopic property is registered, if the system is isolated from its
surroundings.
An isolated system always reaches in course of time a state of thermodynamic
equilibrium and can never depart from it spontaneously.
Therefore, there can be no spontaneous change in any macroscopic property if
the system exists in an equilibrium state. Thermodynamics studies mainly the
properties of physical systems that are found in equilibrium states.
A system will be in a state of thermodynamic equilibrium, if the conditions for
the following three types of equilibrium are satisfied:
(a) Mechanical equilibrium
(b) Chemical equilibrium
(c) Thermal equilibrium
In the absence of any unbalanced force within the system itself and also between
the system and the surroundings, the system is said to be in a state of mechanical
equilibrium. If an unbalanced force exists, either the system alone or both the
system and the surroundings will undergo a change of state till mechanical
equilibrium is attained.
If there is no chemical reaction or transfer of matter from one part of the system
to another, such as diffusion or solution, the system is said to exist in a state of
chemical equilibrium.
12
Engineering Thermodynamics
When a system existing in mechanical and chemical equilibrium is separated
from its surroundings by a diathermic wall (diathermic means ‘which allows heat to
flow’) and if there is no spontaneous change in any property of the system, the
system is said to exist in a state of thermal equilibrium. When this is not satisfied,
the system will undergo a change of state till thermal equilibrium is restored.
When the conditions for any one of the three types of equilibrium are not
satisfied, a system is said to be in a nonequilibrium state. If the nonequilibrium of
the state is due to an unbalanced force in the interior of a system or between the
system and the surroundings, the pressure varies from one part of the system to
another. There is no single pressure that refers to the system as a whole. Similarly,
if the nonequilibrium is because of the temperature of the system being different
from that of its surroundings, there is a nonuniform temperature distribution set up
within the system and there is no single temperature that stands for the system as a
whole. It can thus be inferred that when the conditions for thermodynamic
equilibrium are not satisfied, the states passed through by a system cannot be
described by thermodynamic properties which represent the system as a whole.
Thermodynamic properties are the macroscopic coordinates defined for, and
significant to, only thermodynamic equilibrium states. Both classical and statistical
thermodynamics study mainly the equilibrium states of a system.
1.11 QUASI-STATIC PROCESS
Let us consider a system of gas contained in a cylinder (Fig. 1.13). The system
initially is in an equilibrium state, represented by the properties p1, v1, t1. The
weight on the piston just balances the upward force exerted by the gas. If the weight
is removed, there will be an unbalanced force between the system and the
surroundings, and under gas pressure, the piston will move up till it hits the stops.
The system again comes to an equilibrium state, being described by the properties
p2 , v2 , t 2. But the intermediate states passed through by the system are
nonequilibrium states which cannot be described by thermodynamic coordinates.
Stops
Final state
Weight
W
Piston
Initial state
p1 Gas
v1
t1
System boundary
Fig. 1.13 Transition between Two Equilibrium States by an Unbalanced Force
13
Introduction
Figure 1.14 shows points 1 and 2 as the initial and final equilibrium states joined by
a dotted line, which has got no meaning otherwise. Now if the single weight on the
piston is made up of many very small pieces of weights (Fig. 1.15), and these
weights are removed one by one very slowly from the top of the piston, at any
instant of the upward travel of the piston, if the gas system is isolated, the departure
of the state of the system from the thermodynamic equilibrium state will be
infinitesimally small. So every state passed through by the system will be an
Stops
1
p
p1
Weights
Piston
2
p2
v1
Fig. 1.14
v
System
boundary
p1, v1, t1
v2
Gas
Plot Representing the
Transition between Two
Equilibrium States
Fig. 1.15
Infinitely Slow Transition
of a System by Infinitesimal Force
equilibrium state. Such a process, which is but a locus of all the equilibrium points
passed through by the system, is known as a quasi-static process (Fig. 1.16), ‘quasi’
meaning ‘almost’. Infinite slowness is the characteristic feature of a quasi-static
process. A quasi-static process is thus a succession of equilibrium states. A quasistatic process is also called a reversible process.
PURE SUBSTANCE
A pure substance is defined as one
that is homogeneous and invariable
in chemical composition throughout
its mass. The relative proportions of
the chemical elements constituting
the substance are also constant. Atmospheric air, steam-water mixture and
combustion products of a fuel are
regarded as pure substances. But the
mixture of air and liquid air is not a
pure substance, since the relative
proportions of oxygen and nitrogen
differ in the gas and liquid phases in
equilibrium.
1
Equilibrium states
p
1.12
Quasi-static process
2
v
Fig. 1.16 A Quasi-Static Process
14
Engineering Thermodynamics
The state of a pure substance of given mass can be fixed by specifying two
independent intensive properties, provided the system is in equilibrium. This is
known as the ‘two-property rule’. The state can thus be represented as a point on
thermodynamic property diagrams. Once any two properties of a pure substance
are known, other properties can be determined from the available thermodynamic
relations.
1.13
CONCEPT OF CONTINUUM
From the macroscopic viewpoint, we are always concerned with volumes which
are very large compared to molecular dimensions. Even a very small volume of a
system is assumed to contain a large number of molecules so that statistical
averaging is meaningful and a property value can be assigned to it. Disregarding
the behaviour of individual molecules, matter is here treated as continuous. Let us
consider the mass d m in a volume dV surrounding the point P (Fig. 1.17). The ratio
dm/dV is the average mass density of the system within the volume dV. We suppose
that at first dV is rather large, and is subsequently shrunk about the point P. If we
plot dm/dV against dV, the average density tends to approach an asymptote as dV
increases (Fig. 1.18). However, when dV becomes so small as to contain relatively
few molecules, the average density fluctuates substantially with time as molecules
pass into and out of the volume in random motion, and so it is impossible to speak
of a definite value of dm/dV. The smallest volume which may be regarded as
continuous is dV¢. The density r of the system at a point is thus defined as
Domain of
continuum
dm
dV
Domain of
Molecular effects
r
dm
P
System
dV
d V¢
dV
Fig. 1.17
Fig.1.18
r =
lim
dm
dV Æ dV ¢ dV
Definition of the Macroscopic
Property, Density
(1.1)
15
Introduction
Similarly, the fluid velocity at a point P is defined as the instantaneous velocity of
the centre of gravity of the smallest continuous volume dV¢.
The concept of continuum loses validity when the mean free path of the
molecules approaches the order of magnitude of the dimensions of the vessel, as,
for instance, in highly rarefied gases encountered in high vacuum technology, in
rocket flights at high altitudes and in electron tubes. In most engineering
applications, however, the assumption of a continuum is valid and convenient, and
goes hand in hand with the macroscopic point of view.
1.14 THERMOSTATICS
The science of thermodynamics deals with systems existing in thermodynamic
equilibrium states which are specified by properties. Infinitely slow quasi-static
processes executed by systems are only meaningful in thermodynamic plots. The
name ‘thermodynamics’ is thus said to be a misnomer, since it does not deal with
the dynamics of heat, which is nonquasi-static. The name ‘thermostatics’ then
seems to be more appropriate. However, most of the real processes are dynamic
and nonquasi-static, although the initial and final states of the system might be in
equilibrium. Such processes can be successfully dealt with by the subject. Hence,
the term ‘thermodynamics’ is not inappropriate.
1.15 UNITS AND DIMENSIONS
In the present text, the SI (System International) system of units has been used. The
basic units in this system are given in Table 1.1.
Table 1.1
System: Basic Units
Quantity
Unit
Length (L)
Mass (M)
Time (t)
Amount of substance
Temperature (T)
Electric current
Luminous intensity
Plane angle
Solid angle
Metre
Kilogram
Second
Mole
Kelvin
Ampere
Candela
Radian
Steradian
Symbol
m
kg
s
mol
K
A
cd
rad
sr
The dimensions of all other quantities are derived from these basic units which
are given in Table 1.2.
16
Engineering Thermodynamics
SI System: Derived Units
Table 1.2
Quantity
Unit
Force (F)
Energy E)
Power
Pressure
Frequency
Electric charge
Electric potential
Capacitance
Electrical resistance
Magnetic flux
Magnetic flux density
Inductance
Newton
Joule
Watt
Pascal
Hertz
Coulomb
Volt
Farad
Ohm
Weber
Tesla
Henry
Symbol
Alternative unit
N
J
W
Pa
Hz
C
V
F
W
Wb
T
H
In basic units
kg m/s2
kg m2/s2
kg m2/s3
kg/(ms2)
s –1
As
kg m2/(s3 A)
s4 A2/(kg m2)
kg m2/(s3 A2)
kg m2/(s2 A)
kg/(s2 A)
kg m2/(s2 A2)
Nm
J/s
N/m2
W/A = J/C
C/V
V/A
Vs
Wb/m2
Wb/A
It is often convenient and desirable to use multiples of various units, the standard list of which is given in Table 1.3.
Table 1.3
SI System: Standard Multipliers
Factor
1012
109
106
103
1.15.1
Prefix
Factor
Prefix
tera, T
giga, G
mega, M
kilo, k
10–3
10–6
10–9
10–12
milli, m
micro, m
nano, n
pico, p
Force
The force acting on a body is defined by Newton’s second law of motion. The unit
of force is the newton (N). A force of one newton produces an acceleration of
1 ms–2 when applied to a mass of 1 kg.
1 N = 1 kg m/s2
The weight of a body (W) is the force with which the body is attracted to the
centre of the earth. It is the product of its mass (m) and the local gravitational
acceleration (g), i.e.
W = mg
The value of g at sea level is 9.80665 m/s2. The mass of a substance remains
constant with elevation, but its weight varies with elevation.
1.15.2
dA
Pressure
Pressure is the normal force exerted by a system
against unit area of the bounding surface. If dA is
a small area and dA¢ is the smallest area from continuum consideration, and dFn is the component
of force normal to dA (Fig. 1.19), the pressure p
at a point on the wall is defined as
Fn
System
Fig. 1.19
Definition of
Pressure
17
Introduction
p = lim
d Fn
dA Æ dA¢ d A
The pressure p at a point in a fluid in equilibrium is the same in all directions.
The unit for pressure in the SI system is the pascal (Pa), which is the force of
one newton acting on an area of 1 m2.
1 Pa = 1 N/m2
The unit of pascal is very small. Very often kilo-pascal (kPa) or mega-pascal (MPa)
is used.
Two other units, not within the SI system of units, continue to be widely used.
These are the bar, where
1 bar =105 Pa = 100 kPa = 0.1 MPa
and the standard atmosphere, where
1 atm = 101.325 kPa
= 1.01325 bar
Most instruments indicate pressure relative to the atmospheric pressure, whereas
the pressure of a system is its pressure above zero, or relative to a perfect vacuum
(Fig. 1.20). The pressure relative to the atmosphere is called gauge pressure. The
pressure relative to a perfect vacuum is called absolute pressure.
Absolute pressure = Gauge pressure + Atmospheric pressure
p > 1 atm
Positive gauge pressure
Negative gauge pressure or vacuum
pressure below
atmospheric
p < 1 atm
Absolute
Local atmospheric
pressure
Absolute pressure
above atmosphere
p = 1 atm
p=0
Zero pressure
perfect vacuum
Fig. 1.20 Relationship between Pressures
When the pressure in a system is less than atmospheric pressure, the gauge pressure becomes negative, but is frequently designated by a positive number and called
vacuum. For example, 16 cm vacuum will be
76 – 16
¥ 1.013 = 0.80 bar
76
Thus,
pabs = patm – pvac
18
Engineering Thermodynamics
Figure 1.21 shows a few pressure measuring devices. Figure (a) shows the
Bourdon gauge which measures the difference between the system pressure inside
the tube and atmospheric pressure. It relies on the deformation of a bent hollow
tube of suitable material which, when subjected to the pressure to be measured on
the inside (and atmospheric pressure on the outside), tends to unbend. This moves
a pointer through a suitable gear-and lever mechanism against a calibrated scale.
Figure (b) shows an open U-tube indicating gauge pressure, and Fig. (c) shows an
open U-tube indicating vacuum. Figure (d) shows a closed U-tube indicating
absolute pressure. If p is atmospheric pressure, this is a barometer. These are called
U-tube manometers.
patm
patm
p
Z
Hg
p
(a)
(b)
patm
Evacuated
p
p
Z
Z
Hg
Hg
(c)
Fig. 1.21
(d)
Pressure Gauges (a) Bourdon Gauge (b) Open U-tube Indicating
Gauge Pressure (c) Open U-tube Indicating Vacuum (d) Closed
U-tube Indicating Absolute Pressure
19
Introduction
If Z is the difference in the heights of the fluid columns in the two limbs of the Utube (Fig. (b) and Fig. (c)], r the density of the fluid and g the acceleration due to
gravity, then from the elementary principle of hydrostatics, the gauge pressure pg is
given by
LM
N
OP
Q
kg m
= Zrg N/m2
m 3 s2
If the fluid is mercury having r = 13,616 kg/m3, one metre head of mercury
column is equivalent to a pressure of 1.3366 bar, as shown below
pg = Zrg m ◊
1 mHg = Z r g = 1 ¥ 13616 ¥ 9.81
= 1.3366 ¥ 105 N/m2 = 1.3366 bar
The manometer is a sensitive, accurate and simple device, but it is limited to
fairly small pressure differentials and, because of the inertia and friction of the
liquid, is not suitable for fluctuating pressures, unless the rate of pressure change is
small. A diaphragm-type pressure transducer along with a cathode ray oscilloscope
can be used to measure rapidly fluctuating pressures. The unit of 1 mm Hg pressure
is called torr, so that
1 mmHg = 1 torr = 133 Pa
Figure 1.22 shows a typical U-tube manometer, one end of which is connected to
the vessel the pressure of which is to be measured and the other end is open to
atmosphere exerting pressure patm . The manometric fluid may be mercury, water,
alcohol, oil, etc. The pressure along the horizontal line AB is the same in both the
limbs of the manometer, so that
p + r1 gz1 = patm + r2 gz2
If the vessel contains a gas of density r1 and since the density of the manometric
fluid, (say Hg), r2 >> r1,
p – patm = r2 gz 2 = pgauge
1.15.3
patm
Specific Volume and
Density
Volume (V) is the space occupied by a
substance and is measured in m3. The
specific volume (v) of a substance is
defined as the volume per unit mass and is
measured in m3/kg. From continuum
consideration the specific volume at a
point is defined as
dV
v = lim
dV Æ dV ¢ d m
where dV ¢ is the smallest volume for which
the system can be considered a continuum.
Vessel
r2
z2
p
r1
z1
A
Fig. 1.22
B
Pressure Measurement by a
Manometer
20
Engineering Thermodynamics
Density (r) is the mass per unit volume of a substance, which has been discussed
earlier, and is given in kg/m3.
m
r=
v
In addition to m3, another commonly used unit of volume is the litre (l).
1 l = 10–3 m3
The specific volume or density may be given either on the basis of mass or in
respect of mole. A mole of a substance has a mass numerically equally to the
molecular weight of the substance. One g mol of oxygen has a mass of 32 g and
1 kg mol (or kmol) of nitrogen has a mass of 28 kg. The symbol v is used for molar
specific volume (m3/kmol).
1.15.4
Energy
Energy is the capacity to exert a force through a distance, and manifests itself in
various forms. Engineering processes involve the conversion of energy from one
form to another, the transfer of energy from place to place, and the storage of
energy in various forms, utilizing a working substance.
The unit of energy in the SI system is Nm or J (joule). The energy per unit mass
is the specific energy, the unit of which is J/kg.
1.15.5
Power
The rate of energy transfer or storage is called power. The unit of power is watt
(W), kilowatt (kW) or megawatt (MW).
1 W = 1 J/s = 1 Nm/s
1 kW = 1000 W
Solved Examples
Example 1.1 The pressure of gas in a pipe line is measured with a mercury
manometer having one limb open to the atmosphere (Fig. 1.23). If the difference in
the height of mercury in the two limbs is 562 mm, calculate the gas pressure. The
barometer reads 761 mm Hg, the acceleration due to gravity is 9.79 m/s2, and the
density of mercury is 13,640 kg/m3.
Solution
p
At the plane AB, we have
p = p0 + r g z
p0
Now
p0 = r g z0
z
A
B
where z0 is the barometric height, r the density
of mercury and p0 the atmospheric pressure.
Fig. 1.23
Introduction
21
Therefore
p = r g (z + z0 )
= 13,640 kg/m3 ¥ 9.79 m/s2 (0.562 + 0.761) m
= 177 ¥ 10 3 N/m2 = 177 kPa = 1.77 bar = 1.746 atm
Example 1.2 A turbine is supplied with steam at a gauge pressure of 1.4 MPa.
After expansion in the turbine the steam flows into a condenser which is maintained at a vacuum of 710 mmHg. The barometric pressure is 772 mmHg.
Express the inlet and exhaust steam pressures in pascals (absolute). Take the density of mercury as 13.6 ¥ 103 kg/m3.
Solution
The atmospheric pressure p0
= rgz0 = 13.6 ¥ 103 kg/m3 ¥ 9.81 m/s2 ¥ 0.772 m
= 1.03 ¥ 105 Pa
Inlet steam pressure
= [(1.4 ¥ 10 6) + (1.03 ¥ 105)] Pa
= 15.03 ¥ 105 Pa = 1.503 MPa
Condenser pressure
= (0.772 – 0.710) m ¥ 9.81 m/s2 ¥ 13.6 ¥ 103 kg/m3
= 0.827 ¥ 104 Pa = 8.27 kPa
Example 1.3 Convert the following readings of pressure to kPa, assuming that
the barometer reads 760 mm of Hg.
(a) 40 cmHg vacuum
(b) 90 cmHg gauge
(c) 1.2 m of H2O gauge
Solution (a)
pvacuum = hrg = (40 ¥ 10 –2) ¥ (13.6 ¥ 10 3) ¥ 9.8
= 53.31 ¥ 10 3 N/m2 = 53.31 kPa
pabsolute = patm – pvac
= (760 – 400) ¥ 9.8 ¥ 13.6 ¥ 10 3
= 48 ¥ 103 N/m2 = 48 kPa
Also,
(b)
pabs = 101.325 – 53.31 = 48.015 kPa
pgauge = h r g
= (90 ¥ 10–2) ¥ (13.6 ¥ 10 3) ¥ 9.8
= 120 ¥ 10 3 N/m2 = 120 kPa
Since
patm = 760 mm Hg = 101.325 kPa
pabs = pgauge + patm
= 120 + 101.325 = 221.325 kPa
22
Engineering Thermodynamics
pgauge = h r g = 1.2 m ¥ 1000
(c)
kg
m
¥ 9.81 2
m3
s
= 11.772 kPa
\
pabs = 11.772 + 101.325 = 113.097 kPa
dh
Example 1.4 The pressure and specific volume of the atmosphere are related
according to the equation pv1.4 = 2.5 ¥ 105, where p is in N/m2 and v in m3/kg.
What depth of atmosphere is required to produce a pressure of 1.033 bar at the
earth’s surface? Assume g = 9.81 m/s2.
p
Solution Let H be the depth of the
Atmospheric
atmosphere required to produce the
column of area A
pressure of 1.033 bar at the earth’s
surface. Considering an element of
length dh (Fig. 1.24), by force balance:
p + dp
W
A (p + dp) = mg + pA
pA + Adp = rAdhg + pA
or,
dh =
H
or,
1
vdp
g
z z
dh =
Earth
Fig. 1.24
FG 2.5 ¥ 10 IJ
gH
p
K
1.033 ¥ 105 1
0
0
H=
5
1
1. 4
1
(2.5 ¥ 10 5)0.714
g
dp
z
1. 033 ¥ 10 5
p–0.714 dp
0
5 0. 286
d1.033 ¥ 10 i
= 731.274
0.286
= 69,462 m = 69.462 km
Example 1.5 Acceleration is sometimes measured in g’s, or multiples of the
standard acceleration of gravity. Determine the net upward force that an astronaut
whose mass is 68 kg experiences if the acceleration on lift-off is 10 g’s.
Solution
Net vertical force
m =� 68� kg
F
=
F = mg = ma
a =� 10� g's
net
a = 10 ¥ 9.806 = 98.06 m/s2
F = 68 ¥ 98.06 = 6668 N Ans.
F
Summary
A certain quantity of matter or a region in space upon which attention is focused
in the analysis of a problem is called a system. Everything external to the system
is called the surroundings which is separated from the system by a boundary.
The system and its surroundings together is called a universe.
Introduction
23
Thermodynamics studies the energy interactions like work and heat transfer and
also mass transfer between a system and the surroundings and how these
interactions affect the properties of the system. It is an empirical science drawing
its material from certain observed facts of nature formulated into thermodynamic
laws. Only the macroscopic behaviour of the system is considered, without taking
into account the events occurring at the molecular or microscopic level.
Properties which are the coordinates to describe the state of a system can be
intensive, which are independent of mass, like pressure and temperature.
Extensive properties like volume and energy depend on the mass. Specific
extensive properties are intensive properties.
Systems can be of three types: a closed system can have only energy
interactions and no mass transfer. An open system can have both mass and energy
transfer across its boundary. An isolated system is one in which there is no
interaction, either mass or energy transfer, across the boundary.
Any operation in which one or more of the properties changes is called a change
of state. When the path of the change of state is specified, it is called a process.
A thermodynamic cycle is a series of state changes such that the final state is
identical with the initial state.
A system is said to exist in a state of thermodynamic equilibrium when no
change in any macroscopic property is registered, if the system is isolated from
its surroundings. If a system satisfies three types of equilibrium, viz, mechanical,
chemical and thermal, it is said to be in thermodynamic equilibrium. A quasistatic process is a succession of equilibrium states, occurring infinitely slowly,
so that the process takes infinite time to complete it across a finite gradient.
A pure substance is one that is homogeneous and invariable in chemical
composition throughout its mass like a mixture of ice and water. The twoproperty rule holds that the state of a pure substance of given mass can be fixed
by specifying two independent intensive properties.
Review Questions
1.1 What do you understand by macroscopic and microscopic viewpoints?
1.2 Is thermodynamics a misnomer for the subject?
1.3 How does the subject of thermodynamics differ from the concept of heat
transfer?
1.4 What is the scope of classical thermodynamics?
1.5 What is a thermodynamic system?
1.6 What is the difference between a closed system and an open system?
1.7 An open system defined for a fixed region and a control volume are
synonymous. Explain.
1.8 Define an isolated sysetm.
1.9 Distinguish between the terms ‘change of state’, ‘path’, and ‘process’.
1.10 What is a thermodynamic cycle?
24
Engineering Thermodynamics
1.11
1.12
1.13
1.14
1.15
1.16
What are intensive and extensive properties?
What do you mean by homogeneous and heterogeneous systems?
Explain what you understand by thermodynamic equilibrium.
Explain mechanical, chemical and thermal equilibrium.
What is a quasi-static process? What is its characteristic feature?
What is the concept of continuum? How will you define density and pressure
using this concept?
1.17 What is vacuum? How can it be measured?
1.18 What is a pressure transducer?
Problems
1.1 A pump discharges a liquid into a drum at the rate of 0.032 m3/s. The drum,
1.50 m in diameter and 4.20 m in length, can hold 3000 kg of the liquid. Find the
density of the liquid and the mass flow rate of the liquid handled by the pump.
Ans. 12.934 kg/s
1.2 The acceleration of gravity is given as a function of elevation above sea level by
g = 980.6 – 3.086 ¥ 10 –6 H
where g is in cm/s2, and H is in cm. If an aeroplane weighs 90,000 N at sea level,
what is the gravity force upon it at 10,000 m elevation? What is the percentage
difference from the sea-level weight?
Ans. 89,716.4 N, 0.315%
1.3 Prove that the weight of a body at an elevation H above sea-level is given by
W=
FG
H
mg
d
g0 d + 2 H
IJ
K
2
where d is the diameter of the earth.
1.4 The first artificial earth satellite is reported to have encircled the earth at a speed
of 28,840 km/h and its maximum height above the earth’s surface was stated to
be 916 km. Taking the mean diameter of the earth to be 12,680 km, and assuming
the orbit to be circular, evaluate the value of the gravitational acceleration at this
height.
The mass of the satellite is reported to have been 86 kg at sea-level. Estimate
the gravitational force acting on the satellite at the operational altitude.
Ans. 8.9 m/s2; 765 N
1.5 Convert the following readings of pressure to kPa, assuming that the
barometer reads 760 mmHg:
(a) 90 cmHg gauge, (b) 40 cmHg vacuum, (c) 1.2 m H 2 O gauge,
(d) 3.1 bar.
1.6 A 30 m high vertical column of a fluid of density 1878 kg/m3 exists in a place
where g = 9.65 m/s2. What is the pressure at the base of the column.
Ans. 544 kPa
1.7 Assume that the pressure p and the specific volume v of the atmosphere are
related according to the equation pv1.4 = 2.3 ¥ 105, where p is in N/m2 abs and
25
Introduction
3 cm
50 cm
v is in m3/kg. The acceleration due to gravity is constant at 9.81 m/s2. What is
the depth of atmosphere necessary to produce a pressure of 1.0132 bar at the
earth’s surface? Consider the atmosphere as a fluid column.
Ans. 64.8 km
1.8 The pressure of steam flowing in a pipeline is measured with a mercury manomSteam at pressure, p
eter, shown in Fig. 1.25. Some steam
p0
condenses into water. Estimate the steam
pressure in kPa. Take the density of mercury as 13.6 ¥ 103 kg/m3, density of water as 103 kg/m3, the barometer reading
H2O
Hg
as 76.1 cmHg, and g as 9.806 m/s2.
1.9 A vacuum gauge mounted on a condenser
reads 0.66 mHg. What is the absolute
pressure in the condenser in kPa when
the atmospheric pressure is 101.3 kPa?
Ans. 13.3 kPa
Fig. 1.25
2
Temperature
2.1
ZEROTH LAW OF THERMODYNAMICS
The property which distinguishes thermodynamics from other sciences is
temperature. One might say that temperature bears as important a relation to
thermodynamics as force does to statics or velocity does to dynamics. Temperature
is associated with the ability to distinguish hot from cold. When two bodies at
different temperatures are brought into contact, after some time they attain a common
temperature and are then said to exist in thermal equilibrium.
When a body A is in thermal equilibrium with a body B, and also separately
with a body C, then B and C will be in thermal equilibrium with each other.
This is known as the zeroth law of thermodynamics. It is the basis of temperature
measurement.
In order to obtain a quantitative measure of temperature, a reference body is
used, and a certain physical characteristic of this body which changes with
temperature is selected. The changes in the selected characteristic may be taken as
an indication of change in temperature. The selected characteristic is called the
thermometric property, and the reference body which is used in the determination
of temperature is called the thermometer. A very common thermometer consists of a
small amount of mercury in an evacuated capillary tube. In this case the extension of
the mercury in the tube is used as the thermometric property.
There are five different kinds of thermometer, each with its own thermometric
property, as shown in Table 2.1.
2.2 MEASUREMENT OF TEMPERATURE—THE REFERENCE
POINTS
The temperature of a system is a property that determines whether or not a system
is in thermal equilibrium with other systems. If a body is at, say, 70°C, it will be 70°C,
whether measured by a mercury-in-glass thermometer, resistance thermometer or
constant volume gas thermometer. If X is the thermometric property, let us arbitrarily
choose for the temperature common to the thermometer and to all systems in thermal
equilibrium with it the following linear function of X:
27
Temperature
Table 2.1
Thermometers and Thermometric Properties
Thermometer
Thermometric property
1. Constant volume gas thermometer
2. Constant pressure gas thermometer
3. Electrical resistance thermometer
4. Thermocouple
5. Mercury-in-glass thermometer
Pressure
Volume
Resistance
Thermal e.m.f.
Length
Symbol
p
V
R
e
L
q (X) = aX, where a is an arbitrary constant.
If X1 corresponds to q (X1), then X2 will correspond to
q ( X1 )
X1
q (X2) =
that is
◊ X2
q ( X1 )
X1
◊ X2
(2.1)
Two temperatures on the linear X scale are to each other as the ratio of the
corresponding X’s.
2.2.1
Method Used Before 1954
The thermometer is first placed in contact with the system whose temperature
q (X) is to be measured, and then in contact with an arbitrarily chosen standard
system in an easily reproducible state where the temperature is q (X1). Thus
q ( X1 )
q (X)
=
X1
X
(2.2)
Then the thermometer at the temperature q (X) is placed in contact with
another arbitrarily chosen standard system in another easily reproducible state
where the temperature is q (X2). It gives
q (X2 )
q (X )
=
X2
X
=
X1 – X 2
X
(2.3)
From Eqs. (2.2) and (2.3)
q ( X1 ) – q ( X 2 )
q (X)
or
q (X) =
q ( X1 ) – q ( X 2 )
X1 – X 2
◊X
(2.4)
If we assign an arbitrary number of degrees to the temperature interval q (X1)
– q (X2), then q (X) can be calculated from the measurements of X, X1 and X2 .
28
Engineering Thermodynamics
An easily reproducible state of an arbitrarily chosen standard system is called
a fixed point. Before 1954, there were two fixed points: (a) the ice point, the
temperature at which pure ice coexisted in equilibrium with air-saturated water at
one atmosphere pressure, and (b) the steam point, the temperature of equilibrium
between pure water and pure steam at one atmosphere pressure. The temperature
interval, q(X1) – q(X2), between these two fixed points was chosen to be 100 degrees.
The use of two fixed points was found unsatisfactory and later abandoned,
because of (a) the difficulty of achieving equilibrium between pure ice and airsaturated water (since when ice melts, it surrounds itself only with pure water and
prevents intimate contact with air-saturated water), and (b) extreme sensitiveness of
the steam point to the change in pressure.
2.2.2
Method in Use After 1954
Since 1954 only one fixed point has been in use, viz. the triple point of water, the
state at which ice, liquid water and water vapour coexist in equilibrium. The
temperature at which this state exists is arbitrarily assigned the value of 273.16
degrees Kelvin, or 273.16 K (the reason for using Kelvin’s name will be explained
later). Designating the triple point of water by q t, and with Xt being the value of the
thermometric property when the body, whose temperature q is to be measured, is
placed in contact with water at its triple point, it follows that
q t = aXt
qt
273.16
=
Xt
Xt
\
a=
Therefore
q = aX =
or
q = 273.16
273.16
◊X
Xt
X
Xt
(2.5)
The temperature of the triple point of water, which is an easily reproducible state, is
now the standard fixed point of thermometry.
2.3 COMPARISON OF THERMOMETERS
Applying the above principle to the five thermometers listed in Table 2.1, the
temperatures are given as follows:
(a) Constant volume gas thermometer
q (P) = 273.16
p
pt
(b) Constant pressure gas thermometer
q (V) = 273.16
V
Vt
(c) Electric resistance thermometer
q (R) = 273.16
R
Rt
29
Temperature
(d) Thermocouple
q (e) = 273.16
e
et
(e) Liquid-in-glass thermometer
q (L) = 273.16
L
Lt
If the temperature of a given system is measured simultaneously with each of the
five thermometers, it is found that there is considerable difference among the
readings. The smallest variation is, however, observed among different gas
thermometers. That is why a gas is chosen as the standard thermometric substance.
2.4
IDEAL GAS
It has been established from experimental observations that the p – v – T behaviour
of gases at a low pressure is closely given by the following relation
pv = RT
(2.6)
where R is the universal gas constant, 8.3143 J/mol K and v is the molar specific
volume, m3/gmol. (see Sec. 10.3.). Dividing Eq. (2.6) by the molecular weight m,
pv = RT
(2.7)
3
where v is specific volume, in m /kg, and R is the characteristic gas constant.
Substituting R = R /m J/kg K, we get in terms of the total volume V of gas,
PV = n RT
PV = mRT
(2.8)
where n is the number of moles and m is the mass of the gas. Equation (2.8) can be
written for two states of the gas,
p1V1
pV
= 2 2
T1
T2
(2.9)
Equation (2.6), (2.7) or (2.8) is called the ideal gas equation of state. At very low
pressure or density, all gases and vapours approach ideal gas behaviour.
2.5 GAS THERMOMETERS
A schematic diagram of a constant volume gas thermometer is given in Fig. 2.1. A
small amount of gas is enclosed in bulb B which is in communication via the capillary
tube C with one limb of the mercury manometer M. The other limb of the mercury
manometer is open to the atmosphere and can be moved vertically to adjust the
mercury levels so that the mercury just touches lip L of the capillary. The pressure in
the bulb is used as a thermometric property and is given by
p = p0 + rM Zg
where p0 is the atmospheric pressure, rM is the density of mercury.
When the bulb is brought in contact with the system whose temperature is to be
measured, the bulb, in course of time, comes in thermal equilibrium with the system.
30
Engineering Thermodynamics
p0
The gas in the bulb expands, on
being heated, pushing the mercury
downward. The flexible limb of the
manometer is then adjusted so that
the mercury again touches the lip
L. The difference in mercury level Z
is recorded and the pressure p of
the gas in the bulb is estimated.
Since the volume of the trapped
gas is constant, from the ideal gas
equation,
DT =
V
Dp
R
C
Z
L
B
M
Flexible tubing
(2.10)
i.e. the temperature increase is
proportional to the pressure
increase.
Fig. 2.1 Constant Volume Gas Thermometer
In a constant pressure gas thermometer, the mercury levels have to be adjusted
to keep Z constant, and the volume of gas V, which would vary with the temperature
of the system, becomes the thermometric property.
p
(2.11)
DV
R
i.e. the temperature increase is proportional to the observed volume increase. The
constant volume gas thermometer is, however, mostly in use, since it is simpler in
construction and easier to operate.
\
2.6
DT =
IDEAL GAS TEMPERATURE
Let us suppose that the bulb of a constant volume gas thermometer contains an
amount of gas such that when the bulb is surrounded by water at its triple point, the
pressure pt is 1000 mmHg. Keeping the volume V constant, let the following
procedure be conducted:
(a) Surround the bulb with steam condensing at 1 atm, determine the gas pressure
p and calculate
q = 273.16
p
1000
(b) Remove some gas from the bulb so that when it is surrounded by water at its
triple point, the pressure pt is 500 mmHg. Determine the new value of p and
then q for steam condensing at 1 atm.
q = 273.16
p
500
31
Temperature
(c) Continue reducing the amount of gas in the bulb so that pt and p have smaller
and smaller values, e.g. pt having, say, 250 mmHg, 100 mmHg, and so on. At
each value of pt calculate the corresponding q.
(d) Plot q vs. pt and extrapolate the curve to the axis where pt = 0. Read from the
graph
lim q
pt Æ 0
The graph, as shown in Fig. 2.2, indicates that although the readings of a constant
volume gas thermometer depend upon the nature of the gas, all gases indicate the
same temperature as pt is lowered and made to approach zero.
O2
Air
373.15
N2
q (steam)
= 373.15 K
q (K)
H2
0
250
500
1000
p t, mm Hg
Fig. 2.2
Ideal Gas Temperature for Steam Point
A similar series of tests may be conducted with a constant pressure gas
thermometer. The constant pressure may first be taken to be 1000 mmHg, then
500 mmHg, etc. and at each value of p, the volumes of gas V and Vt may be recorded
when the bulb is surrounded by steam condensing at 1 atm and the triple point of
water, respectively. The corresponding value of q may be calculated from
q = 273.16
V
Vt
and q vs. p may be plotted, similar to Fig. 2.2. It is found from the experiments that all
gases indicate the same value of q as p approaches zero.
Since a real gas, as used in the bulb, behaves as an ideal gas as pressure
approaches zero (which would be explained later in Chapter 10), the ideal gas
temperature T is defined by either of the two equations
T = 273.16 lim
p
pt
pt Æ 0
= 273.16 lim
pÆ0
V
Vt
(2.12)
32
Engineering Thermodynamics
where q has been replaced by T to denote this particular temperature scale, the ideal
gas temperature scale.
450
400
Steam point
S
373.15
T, K
350
Ice point
300
i
273.15
250
200
0
400
800
1200
pt, mmHg
Fig. 2.3 Steam-point and Ice-point from Constant Volume Gas Thermometer
If ps and pt are the measured pressures at the steam point and the triple point
respectively, one gets the value of the steam point temperature T s as
ps
pt Æ 0 pt
Ts = 273.16 lim
(2.13)
which is equal to 373.15 K.
Similarly, the temperature Ti at the ice point is
Ti = 273.16 lim
pi
pt Æ 0 pt
(2.14)
which is equal to 273.15 K (Fig. 2.3).
Alternatively, if the ratio ps /pi is plotted against pt with different gases in the
bulb, one gets different curves as in Fig. 2.2. However, when extrapolated to zero
pressure all curves converge, and the ratio ps /pi tends to a constant value giving
p
Ts
= lim s = 1.366099
pt Æ 0 pi
Ti
(2.15)
This value may be considered as a universal constant.
One may now decide to have a certain number of divisions between the steam
point and the ice point, say 100 as in the Kelvin and Celsius scales so that
Ts – Ti = 100
Solving Eqs. (2.15) and (2.16),
Ts = 373.15 K and Ti = 273.15 K
(2.16)
33
Temperature
2.7 CELSIUS TEMPERATURE SCALE
The Celsius temperature scale employs a degree of the same magnitude as that of
the ideal gas scale, but its zero point is shifted, so that the Celsius temperature of the
triple point of water is 0.01 degree Celsius or 0.01°C. If t denotes the Celsius
temperature, then
t = T – 273.15°
Thus the Celsius temperature ts at which steam condenses at 1 atm. pressure
t s = Ts – 273.15°
= 373.15 – 273.15 = 100.00°C
Similar measurements for ice points show this temperature on the Celsius scale to
be 0.00°C. The only Celsius temperature which is fixed by definition is that of the
triple point.
2.8
ELECTRICAL RESISTANCE THERMOMETER
In the resistance thermometer
(Fig. 2.4) the change in resistance
of a metal wire due to its change in
temperature is the thermometric
property. The wire, frequently
platinum, may be incorporated in a
Wheatstone bridge circuit. The
platinum resistance thermometer
measures temperature to a high
degree ofaccuracy and sensitivity,
which makes it suitable as a
standard for the calibration of
other thermometers.
In a restricted range, the
following quadratic equation is
often used
G
Wheatstone
bridge
R
Fig. 2.4 Resistance Thermometer
2
R = R0 (1 + At + Bt )
where R0 is the resistance of the platinum wire when it is surrounded by melting ice
and A and B are constants.
2.9
THERMOCOUPLE
A thermocouple circuit made up from joining two wires A and B made of dissimilar
metals is shown in Fig. 2.5. Due to the Seebeck effect, a net e.m.f. is generated in the
circuit which depends on the difference in temperature between the hot and cold
junctions and is, therefore, a thermometric property of the circuit. This e.m.f. can be
measured by a microvoltmeter to a high degree of accuracy. The choice of metals
depends largely on the temperature range to be investigated, and copper-
34
Engineering Thermodynamics
constantan, chromel –alumel and platinum–platinum–rhodium are typical combinations in use.
Wire A
To potentiometer
Wire B
Test junction
Copper wires
Ice–water mixture
Reference junction
Fig. 2.5
Thermocouple
A thermocouple is calibrated by measuring the thermal e.m.f. at various known
temperatures, the reference junction being kept at 0°C. The results of such
measurements on most thermocouples can usually be represented by a cubic
equation of the form
e = a + bt + ct2 + dt3
where e is the thermal e.m.f. and the constants a, b, c and d are different for each
thermocouple.
The advantage of a thermocouple is that it comes to thermal equilibrium with the
system, whose temperature is to be measured, quite rapidly, because its mass is
small.
2.10
INTERNATIONAL PRACTICAL TEMPERATURE SCALE
An international temperature scale was adopted at the Seventh General Conference
on Weights and Measures held in 1927. It was not to replace the Celsius or ideal gas
scales, but to provide a scale that could be easily and rapidly used to calibrate
scientific and industrial instruments. Slight refinements were incorporated into the
scale in revisions adopted in 1948, 1954, 1960 and 1968. The international practical
scale agrees with the Celsius scale at the defining fixed points listed in Table 2.2.
The temperature interval from the oxygen point to the gold point is divided into
three main parts, as given below.
(a) From 0 to 660°C A platinum resistance thermometer with a platinum wire
whose diameter must lie between 0.05 and 0.20 mm is used, and the temperature is
given by the equation
R = R0 (1 + At + Bt2)
where the constants R0, A, and B are computed by measurements at the ice point,
steam point, and sulphur point.
35
Temperature
Table 2.2
Temperatures of Fixed Points
Temperature °C
Normal boiling point of oxygen
Triple point of water (standard)
Normal boiling point of water
Normal boiling point of sulphur
(Normal melting point of zinc-suggested
as an alternative to the sulphur point)
Normal melting point of antimony
Normal melting point of silver
Normal melting point of gold
– 182.97
+ 0.01
100.00
444.60
419.50
630.50
960.80
1063.00
(b) From –190 to 0°C
The same platinum resistance thermometer is used, and
the temperature is given by
R = R0 [1 + At + Bt2 + C (t – 100) t3]
where R 0, A and B are the same as before, and C is determined from a measurement
at the oxygen point.
(c) From 660 to 1063°C A thermocouple, one wire of which is made of platinum and the other of an alloy of 90% platinum and 10% rhodium, is used with one
junction at 0°C. The temperature is given by the formula
e = a + bt + ct2
where a, b, and c are computed from measurements at the antimony point, silver
point, and gold point. The diameter of each wire of the thermocouple must lie
between 0.35 and 0.65 mm.
An optical method is adopted for measuring temperatures higher than the gold
point. The intensity of radiation of any convenient wavelength is compared with the
intensity of radiation of the same wavelength emitted by a black body at the gold
point. The temperature is then determined with the help of Planck’s law of thermal
radiation.
Solved Examples
Example 2.1 Two mercury-in-glass thermometers are made of identical
materials and are accurately calibrated at 0°C and 100°C. One has a tube of
constant diameter, while the other has a tube of conical bore, ten per cent greater
in diameter at 100°C than at 0°C. Both thermometers have the length between 0
and 100 subdivided uniformly. What will be the straight bore thermometer read in
a place where the conical bore thermometer reads 50°C?
Solution
The volume of mercury in the tube at t°C, Vt, is given by
Vt = V0 [1 + b (t – t0)]
36
Engineering Thermodynamics
where V0 is the volume of mercury at 0°C, b is the coefficient of volume expansion of
mercury, and t0 is the ice point temperature which is 0°C. The volume change of
glass is neglected.
Vt – V0 = b V0 t
Therefore
The temperature t is thus a linear function of volume change of mercury
(Vt – V0).
Therefore
DV 0– 100 = b V0 ◊ 100
DV0– 50 = b V0 ◊ 50
DV0 – 50
\
D V0 –100
=
1
2
i.e. at 50°C, the volume of mercury will be half of that at 100°C, for the straight bore
thermometer [Fig. 2.6(a)].
But if the bore is conical [Fig. 2.6 (b)], mercury will fill up the volume ACDB,
which is less than half of the mercury volume at 100°C, i.e. volume AEFB. Let t be the
true temperature when mercury rises half the length of the conical tube (the apparent
temperature being 50°C). Let EA and FB be extended to meet at G. Let l represent the
length of the thermometers and l¢ the vertical height of the cone ABG, as shown in
the figure. Now,
l¢
1
d
=
=
l + l ¢ 1.1d 1.1
l¢ = 10
l¢
d
=
l ¢ + l /2 CD
and
1.1d
E
100∞C
F
100∞C
50∞C
50∞C
C
D
l
l/2
0∞C
0∞C
A
B
d
l¢
(a)
G
(b)
Fig. 2.6
Temperature
\
CD =
37
10.5
d = 1.05d
10
Again
D V 0–100 = V0 ◊ b ◊ 100
D V0 – t = V0 b t
D V0 – t
D V0 –100
=
t
100
Volume ACDB
t
=
Volume AEFB
100
or
or
1p
1p
(1.05d )2 ¥ 10.5 l – 3 4 d 2 ◊10l t
34
=
1p
1p 2
2
100
(1.1d ) ¥ 11l – d ◊10l
34
34
or
1.05 ¥ 1.05 ¥ 10.5 - 10
t
=
1.1 ¥ 1.1 ¥ 11 - 10
100
\
t=
1.58
¥ 100 = 47.7°C
3.31
Example 2.2 The e.m.f. in a thermocouple with the test junction at t°C on gas
thermometer scale and reference junction at ice point is given by
e = 0.20 t – 5 ¥ 10–4 t2 mV
The millivoltmeter is calibrated at ice and steam points. What will this thermometer
read in a place where the gas thermometer reads 50°C?
Solution
At ice point, when t = 0°C, e = 0 mV
At steam point, when
t = 100°C, e = 0.20 ¥ 100 – 5 ¥ 10–4 ¥ (100)2
= 15 mV
At
t = 50°C, e = 0.20 ¥ 50 – 5 ¥ 10–4 (50)2 = 8.75 mV
When the gas thermometer reads 50°C, the thermocouple will read
100
¥ 8.75, or 58.33°C
15
Example 2.3 A platinum resistance thermometer has a resistance of 2.8 ohm at
0°C and 3.8 ohm at 100°C. Calculate the temperature when the resistance
indicated is 5.8 ohm.
Solution Let R = R0 (1 + at)
where R0 is the resistance at 0°C.
Therefore,
R0 = 2.8 ohm
38
Engineering Thermodynamics
R 100 = 3.8 ohm = 2.8 (1 + a ¥ 100)
Ê 3.8 ˆ
a=Á
- 1 ¥ 10–2 = 0.357 ¥ 10–2
Ë 2.8 ˜¯
\
When
R = 5.8 ohm,
5.8 = 2.8 (1 + 0.357 ¥ 10–2 t)
100
Ê 5.8 ˆ
t=Á
- 1˜ ¥
= 300°C.
Ë 2.8 ¯ 0.357
or,
Summary
When two bodies at two different temperatures are brought into contact, after
some time they attain a common temperature and are then said to exist in
thermal equilibrium.
When a body A is in thermal equilibrium with a body B and also separately
with body C, then B and C will be in thermal equilibrium with each other.
This is known as the zeroth law of thermodynamics. It is the basis of
temperature measurement. A physical characteristic of an arbitrarily chosen
body which changes with change in temperature is called thermometric
property (X) and the reference body is called the thermometer. The common
thermometers are mercury-in-glass, (length of mercury column, L µ t),
resistance (R µ t), thermocouple (e µ t), constant volume gas thermometer
(p µ t) and constant pressure gas thermometer (v µ t). Before 1954, two fixed
points, the ice point and steam point, were used to quantify the temperature of
a system. After 1954, only one fixed point, the triple point of water (0.01°C)
is used and it is the standard fixed point of thermometry. A thermometer scale
is set up by assuming a linear variation of temperature with the reading of
some arbitrarily selected standard thermometer. The most suitable standard
has been found to be the gas thermometer, which is used in laboratories to
determine the reference points at which other thermometers are calibrated.
An absolute temperature scale may be constructed by taking the
temperature directly proportional to the volume of the gas in a constant
pressure gas thermometer. The scale, based upon a gas at zero pressure, is the
physical realisation of a logically formulated absolute thermodynamic
temperature scale, which will be introduced in connection with the Second
Law of Thermodynamics. For ordinary purposes absolute temperatures may
be found from the relation (°C = 273.15 K).
Review Questions
2.1 What is the zeroth law of thermodynamics?
2.2 Define thermometric property.
Temperature
39
2.3 What is a thermometer?
2.4 What is a fixed point?
2.5 How many fixed points were used prior to 1954? What are these?
2.6 What is the standard fixed point in thermometry? Define it.
2.7 Why is a gas chosen as the standard thermometric substance?
2.8 What is an ideal gas?
2.9 What is the difference between the universal gas constant and a characteristic gas constant?
2.10 What is a constant volume gas thermometer? Why is it preferred to a constant
pressure gas thermometer?
2.11 What do you understand by the ideal gas temperature scale?
2.12 How can the ideal gas temperature for the steam point be measured?
2.13 What is the Celsius temperature scale?
2.14 What is the advantage of a thermocouple in temperature measurement?
2.15 How does the resistance thermometer measure temperature?
2.16 What is the need of the international practical temperature scale?
Problems
2.1 The limiting value of the ratio of the pressure of gas at the steam point and at
the triple point of water when the gas is kept at constant volume is found to be
1.36605. What is the ideal gas temperature of the steam point? Ans. 100°C
2.2 In a constant volume gas thermometer the following pairs of pressure readings were taken at the boiling point of water and the boiling point of sulphur,
respectively:
Water b.p.
50.0
100
200
300
Sulphur b.p.
96.4
193
387
582
The numbers are the gas pressures, mm Hg, each pair being taken with the
same amount of gas in the thermometer, but the successive pairs being taken
with different amounts of gas in the thermometer. Plot the ratio of Sb.p.:H2Ob.p.
against the reading at the water boiling point, and extrapolate the plot to zero
pressure at the water boiling point. This gives the ratio of Sb.p.:H2Ob.p. on a gas
thermometer operating at zero gas pressure, i.e., an ideal gas thermometer.
What is the boiling point of sulphur on the gas scale, from your plot?
Ans. 445°C
2.3 The resistance of a platinum wire is found to be 11,000 ohms at the ice point,
15.247 ohms at the steam point, and 28.887 ohms at the sulphur point. Find the
constants A and B in the equation
R = R0 (1 + At + Bt 2)
and plot R against t in the range 0 to 660°C.
2.4 When the reference junction of a thermocouple is kept at the ice point and the
test junction is at the Celsius temperature t, and e.m.f. e of the thermocouple is
given by the equation
e = at + bt2
where a = 0.20 mV/deg, and b = – 5.0 ¥ 10–4 mV/deg2
(a) Compute the e.m.f. when t = – 100°C, 200°C, 400°C, and 500°C, and draw
graph of e against t in this range.
40
Engineering Thermodynamics
(b) Suppose the e.m.f. e is taken as a thermometric property and that a temperature scale t* is defined by the linear equation.
t* = a¢ e + b¢
and that t* = 0 at the ice point and t* = 100 at the steam point. Find the
numerical values of a¢ and b¢ and draw a graph of e against t*.
(c) Find the values of t* when t = – 100°C, 200°C, 400°C, and 500°C, and draw
a graph of t* against t.
(d) Compare the Celsius scale with the t* scale.
2.5 The temperature t on a thermometric scale is defined in terms of a property K
by the relation
t = a ln K + b
where a and b are constants.
The values of K are found to be 1.83 and 6.78 at the ice point and the steam
point, the temperatures of which are assigned the numbers 0 and 100 respectively. Determine the temperature corresponding to a reading of K equal to
2.42 on the thermometer.
Ans. 21.346°C
2.6 The resistance of the windings in a certain motor is found to be 80 ohms at
room temperature (25°C). When operating at full load under steady state conditions, the motor is switched off and the resistance of the windings, immediately measured again, is found to be 93 ohms. The windings are made of
copper whose resistance at temperature t°C is given by
R t = R 0 [1 + 0.00393 t]
where R 0 is the resistance at 0°C. Find the temperature attained by the coil
during full load.
Ans. 70.41°C
2.7 A new scale N of temperature is divided in such a way that the freezing point
of ice is 100°N and the boiling point is 400°N. What is the temperature reading
on this new scale when the temperature is 150°C? At what temperature both
the Celsius and the new temperature scale reading would be the same?
Ans. 550°N, – 50°C.
2.8 A platinum wire is used as a resistance thermometer. The wire resistance was
found to be 10 ohm and 16 ohm at ice point and steam point respectively, and
30 ohm at sulphur boiling point of 444.6°C. Find the resistance of the wire at
500°C, if the resistance varies with temperature by the relation.
R = R 0 (1 + a t + bt2)
Ans. 31.3 ohm
3
Work and Heat Transfer
A closed system and its surroundings can interact in two ways: (a) by work transfer,
and (b) by heat transfer. These may be called energy interactions and these bring
about changes in the properties of the system. Thermodynamics mainly studies these
energy interactions and the associated property changes of the system.
3.1
WORK TRANSFER
Work is one of the basic modes of energy transfer. In mechanics the action of a
force on a moving body is identified as work. A force is a means of transmitting an
effect from one body to another. But a force itself never produces a physical effect
except when coupled with motion and hence it is not a form of energy. An effect
such as the raising of a weight through a certain distance can be performed by using
a small force through a large distance or a large force through a small distance. The
product of force and distance is the same to accomplish the same effect. In
mechanics work is defined as:
The work is done by a force as it acts upon a body moving in the direction of the
force.
The action of a force through a distance (or of a torque through an angle) is
called mechanical work since other forms of work can be identified, as discussed
later. The product of the force and the distance moved parallel to the force is the
magnitude of mechanical work.
In thermodynamics, work transfer is considered as occurring between the system
and the surroundings. Work is said to be done by a system if the sole effect on things
external to the system can be reduced to the raising of a weight. The weight may
not actually be raised, but the net effect external to the system would be the raising
of a weight. Let us consider the battery and the motor in Fig. 3.1 as a system. The
motor is driving a fan. The system is doing work upon the surroundings. When the
fan is replaced by a pulley and a weight, as shown in Fig. 3.2, the weight may be
raised with the pulley driven by the motor. The sole effect on things external to the
system is then the raising of a weight.
42
Engineering Thermodynamics
Pulley
Motor
Motor
Fan
W
+
+
Surroundings
Battery
Battery
Weight
System boundary
Fig. 3.1
System boundary
Battery-motor System
Driving a Fan
Fig. 3.2
Work Transfer
from a System
When work is done by a system, it is arbitrarily taken to be positive, and when
work is done on a system, it is taken to be negative (Fig. 3.3). The symbol W is used
for work transfer.
W
System
W
System
Surroundings
Surroundings
(a) W is positive
(b) W is negative
Fig. 3.3 Work Interaction between a System and the Surroundings
The unit of work is N.m or Joule [1 Nm = 1 Joule]. The rate at which work is
done by, or upon, the system is known as power. The unit of power is J/s or watt.
Work is one of the forms in which a system and its surroundings can interact with
each other. There are various types of work transfers which can get involved
between them.
3.2 pdV-WORK OR DISPLACEMENT WORK
Let the gas in the cylinder (Fig. 3.4) be a system having initially the pressure p1 and
volume V1. The system is in thermodynamic equilibrium, the state of which is
described by the coordinates p1, V1. The piston is the
Gas system
only boundary which moves due to gas pressure. Let
the piston move out to a new final position 2, which
p2
is also a thermodynamic equilibrium state specified
p1
by pressure p2 and volume V2. At any intermediate
V1
V2
point in the travel of the piston, let the pressure be p
and the volume V. This must also be an equilibrium
1
2
state, since macroscopic properties p and V are
Fig.
3.4
pdV
Work
43
Work and Heat Transfer
significant only for equilibrium states. When the piston moves an infinitesimal
distance dl, and if ‘a’ be the area of the piston, the force F acting on the piston F =
p.a. and the infinitesimal amount of work done by the gas on the piston
dW = F ◊ dl = padl = pdV
(3.1)
_
where dV = adl = infinitesimal displacement volume. The differential sign in d W
with the line drawn at the top of it will be explained later.
When the piston moves out from position 1 to position 2 with the volume
changing from V1 to V2, the amount of work W done by the system will be
z
V2
pdV
V1
The magnitude of the work done is
given by the area under the path 1–2, as
shown in Fig. 3.5. Since p is at all times
a thermodynamic coordinate, all the
states passed through by the system as
the volume changes from V1 to V2 must
be equilibrium states, and the path 1–2
must be quasi-static. The piston moves
infinitely slowly so that every state
passed through is an equilibrium state.
The integration Ú pdV can be performed
only on a quasi-static path.
3.2.1
p1
1
Quasi-static
process
p
W1–2 =
Work
transfer
p
p2
V1
2
dV
V2
V
Fig. 3.5
Quasi-Static pdV Work
Path Function and Point Function
p
With reference to Fig. 3.6, it is possible to take a system from state 1 to state 2
along many quasi-static paths, such as A, B or C. Since the area under each curve
represents the work for each process, the amount of work involved in each case is
not a function of the end states of the process, and it depends on the path the system
follows in going from state 1 to state 2. For this reason, work is called a path
_
function, and d W is an inexact or imperfect differential.
Thermodynamic properties are point
functions, since for a given state, there is
1
a definite value for each property. The
p1
change in a thermodynamic property of a
system in a change of state is independent
C
of the path the system follows during the
B
A
change of state, and depends only on the
initial and final states of the system. The
p2
2
differentials of point functions are exact
or perfect differentials, and the inteV1
V2
gration is simply.
z
V2
V1
V
dV = V2 – V1
Fig. 3.6
Work—A Path Function
44
Engineering Thermodynamics
The change in volume thus depends only on the end states of the system
irrespective of the path the system follows.
On the other hand, work done in a quasi-static process between two given states
depends on the path followed.
z
z
2 _
1
Rather,
d W π W2 – W1
2 _
1
d W = W1–2
or
1 W2
_
To distinguish an inexact differential d W from an exact differential dV or dp the
differential sign is being cut by a line at its top.
From Eq. (3.1),
1 _
dV = d W
(3.2)
p
_
Here, 1/p is called the integrating factor. Therefore, an inexact differential d W
when multiplied by an integrating factor 1/p becomes an exact differential dV.
For a cyclic process, the initial and final states of the system are the same, and
hence, the change in any property is zero, i.e.
where the symbol
z
z
z
dV = 0,
z
dp = 0,
dT = 0
(3.3)
denotes the cyclic integral for the closed path. Therefore, the
cyclic integral of a property is always zero.
3.2.2 pdV-Work in Various Quasi-Static Processes
(a) Constant pressure process (Fig. 3.7) (isobaric or isopiestic process)
W1–2 =
z
z
V2
V1
pdV = p(V2 – V1 )
(3.4)
(b) Constant volume process (Fig. 3.8) (isochoric process)
W1–2 =
(3.5)
2
p1
1
p2
2
p
p
1
pdV = 0
W1-2
V1
Fig. 3.7
V
V2
Constant Pressure
Process
V
Fig. 3.8
Constant Volume
Process
45
Work and Heat Transfer
(c) Process in which pV = C (Fig. 3.9)
\
W1–2 =
z
V2
pdV
pV = p1 V1 = C
V1
p=
W1–2 = p1V1
z
V2 dV
V1
= p1V1 ln
V
bp V g
1 1
V
V2
V1
p1
p2
(d) Process in which pV n = C, where n is a constant (Fig. 3.10).
= p1V1 ln
(3.6)
pV n = p1 V n1 = p2 V 2n = C
\
p=
\
W1–2 =
dp V i
n
1 1
n
V
z
V2
pdV =
V1
V2 p V n
1 1
◊ dV = (p1 V n1)
n
V
z
1
V
1
p
p
Fig. 3.9
V1
2
2
W1-2
V1
V2
pV n = C
(Quasistatic)
n=
n= 1
n
2
=
3
2
2
2
n=µ
p2
–n + 1
n=0
p1
pV = C
(Quasistatic)
LM V OP
N - n + 1Q
V2
V
Process in Which
pV = Constant
=
V
Fig. 3.10
Process in which
pV n = Constant
p1V1n
p V n ¥ V21 – n - p1V1n ¥ V11- n
(V21–n – V11–n ) = 2 2
1- n
1- n
LM F I
MN GH JK
p V - p2 V2 p1V1
p
=
= 1 1
1- 2
p1
n –1
n –1
n –1/ n
OP
PQ
(3.7)
Similarly, for process in which pvg = c, where g = qo/cv,
g -1 ˘
È
p1V1 Í Ê p2 ˆ g ˙
1W1–2 =
˙
n –1 Í ÁË p1 ˜¯
ÍÎ
˙˚
(3.7a)
46
3.3
Engineering Thermodynamics
INDICATOR DIAGRAM
An indicator diagram is a trace made by a recording pressure gauge, called the
indicator, attached to the cylinder of a reciprocating engine. This represents the
work done in one engine cycle. Figure 3.11 shows a typical engine indicator.
D
Pencil
point
L
C
I
R
Connecting rod
Piston
Crank
P
N
IDC
Cross-head
Piston rod
Fig. 3.11
Crank
shaft
ODC
Stroke
Engine Indicator
The same gas pressure acts on both the engine piston P and the indicator piston I.
The indicator piston is loaded by a spring and it moves in direct proportion to the
change in pressure. The motion of the indicator piston causes a pencil held at the
end of the linkage L to move upon a strip of paper wrapped around drum D. The
drum is rotated about its axis by cord C, which is connected through a reducing
motion R to the piston P of the engine. The surface of drum D moves horizontally
under the pencil while the pencil moves vertically over the surface and a plot of
pressure upon the piston vs. piston travel is obtained.
Before tracing the final indicator diagram, a pressure reference line is recorded
by subjecting the indicator to the atmosphere and tracing a line at a constant pressure
of one atmosphere.
The area of the indicator diagram represents the magnitude of the net work done
by the system in one engine cycle. The area under the path 1–2 represents work
done by the system and the area under the path 2–1 represents work done upon the
system (Fig. 3.12). The area of the diagram, ad , is measured by means of a
planimeter, and the length of the diagram, ld, is also measured. The mean effective
pressure (m.e.p.) pm is defined in the following way
pm =
ad
¥K
ld
where K is the indicator spring constant (N/cm2 ¥ cm travel). Work done in one
engine cycle
= (pm ◊ A) L
47
Work and Heat Transfer
where
A = cross-sectional area of the cylinder
=
L = stroke of piston, or length of cylinder.
Pressure
and
p 2
D , where D is the cylinder diameter
4
1
Represents work done
in one engine cycle
Area, ad
2
Length, Id
1 atm
Piston travel
Volume
Fig. 3.12 Indicator Diagram
Let N be the revolutions per minute (r.p.m.) of the crankshaft. In a two-stroke
cycle, the engine cycle is completed in two strokes of the piston or in one revolution
of the crankshaft. In a four-stroke cycle, the engine cycle is completed in four
strokes of the piston or two revolutions of the crankshaft.
For a two-stroke engine, work done in one minute = pm ALN, and for a fourstroke engine, work done in one minute = pm ALN/2.
The power developed inside the cylinder of the engine is called indicated power
(IP),
\
F
H
pm AL N or
IP =
60
I
K kW
N
n
2
(3.8)
where pm is in kPa and n is the number of cylinders in the engine.
The power available at the crankshaft is always less than this value (IP) due to
friction, etc. and is called the brake power (BP) or shaft power (SP). If w is the
angular velocity of the crankshaft in radian/sec, then
BP = T w
(3.9)
where T is the torque transmitted to the crankshaft in mN.
\
BP =
2p TN
60
(3.10)
where N is the number of revolutions per minute (rpm).
The mechanical efficiency of the engine, hmech, is defined as
hmech =
BP
IP
(3.11)
48
Engineering Thermodynamics
An engine is said to be double-acting, if the working fluid is made to work on
both sides of the piston. Such an engine theoretically develops twice the amount of
work developed in a single-acting engine. Most reciprocating steam engines are
double-acting, and so are many marine diesel engines. Internal combustion engines
for road transport are always single-acting.
3.4
OTHER TYPES OF WORK TRANSFER
There are forms of work other than pdV or displacement work. The following are
the additional types of work transfer which may get involved in system-surroundings
interactions.
(a) Electrical Work When a current flows
through a resistor (Fig. 3.13), taken as a system,
there is work transfer into the system. This is because the current can drive a motor, the motor can
drive a pulley and the pulley can raise a weight.
The current flow, I, in amperes, is given by
I=
I
I
System boundary
Fig. 3.13 Electrical Work
dC
dt
where C is the charge in coulombs and t is time in seconds. Thus dC is the charge
crossing a boundary during time dt. If E is the voltage potential, the work is
_
d W = E ◊ dC
= EI dt
\
W=
z
2
EI dt
(3.12)
1
The electrical power will be
.
W = lim
dW
dt Æ 0 dt
= EI
(3.13)
This is the rate at which work is transferred.
(b) Shaft Work When a shaft, taken
as the system (Fig. 3.14), is rotated by
a motor, there is work transfer into the
system. This is because the shaft can
rotate a pulley which can raise a
weight. If T is the torque applied to the
shaft and dq is the angular displacement of the shaft, the shaft work is
W=
z
System boundary
T
Motor
N
Fig. 3.14
Shaft
Shaft Work
2
1
Tdq
(3.14 a)
Work and Heat Transfer
49
z
(3.15b)
and the shaft power is
W=
2
T
1
dq
= Tw
dt
where w is the angular velocity and T is considered a constant in this case.
(c) Paddle-Wheel Work or Stirring Work As the weight is lowered, and the
paddle wheel turns (Fig. 3.15), there is work transfer into the fluid system which
gets stirred. Since the volume of the system remains constant,
z
pdV = 0. If m is the
mass of the weight lowered through a distance dz and T is the torque transmitted by
the shaft in rotating through an angle dq, the differential work transfer to the fluid is
given by
System
W
Weight
Fig. 3.15 Paddle-Wheel Work
dW = mgdz = Tdq
and the total work transfer is
W=
z
2
1
mgdz =
z
2
1
W¢ dz =
z
2
1
Tdq
(3.15)
where W¢ is the weight lowered.
(d) Flow Work
The flow work,
significant only in a flow process
or an open system, represents the
energy transferred across the system boundary as a result of the energy imparted to the fluid by a
pump, blower or compressor to
make the fluid flow across the control volume. Flow work is analogous to displacement work. Let p
be the fluid pressure in the plane
of the imaginary piston, which
acts in a direction normal to it
(Fig. 3.16). The work done
1 p1, V1, A1
m1
Boundary
Imaginary
Piston
dV
2 p2, V2, A2
m2
Fig.
3.16
Flow
Work
50
Engineering Thermodynamics
on this imaginary piston by the external pressure as the piston moves forward
is given by
_
d Wflow = p dV,
(3.16)
where dV is the volume of fluid element about to enter the system.
_
d Wflow = pv dm
\
where dV = v dm
(3.17)
Therefore, flow work at inlet (Fig. 3.16),
(dWflow)in = p1v1 dm1
(3.18)
Equation (3.18) can also be derived in a slightly different manner. If the normal
pressure p1 is exerted against the area A1, giving a total force (p1 A1) against the
piston, in time dt, this force moves a distance V1dt, where V1 is the velocity of
flow (piston). The work in the time dt is p1 A1 V1 dt, or the work per unit time is p1
A1 V1. Since the flow rate
A V
d m1
w1 = 1 1 =
v1
dt
the work done in time dt becomes
_
( d Wflow)in = p1v1 dm1
Similarly, flow work of the fluid element leaving the system is
_
( d Wflow)out = p2v2 dm2
(3.19)
The flow work per unit mass is thus
_
d Wflow = pv
(3.20)
It is the displacement work done at the moving system boundary.
(e) Work Done in Stretching a Wire
Let us consider a wire as
the system. If the length of the wire in which there is a tension F is changed from L
to L + dL, the infinitestimal amount of work that is done is equal to
_
d W = – F dL
The minus sign is used because a positive value of dL means an expansion of the
wire, for which work must be done on the wire, i.e. negative work. For a finite
change of length,
W=–
z
2
1
F dL
(3.21)
If we limit the problem to within the elastic limit, where E is the modulus of
elasticity, s is the stress, e is the strain, and A is the cross-sectional area, then
F = sA = EeA, since
de =
dL
L
s
=E
e
Work and Heat Transfer
51
_
d W = – F dL = – Ee AL de
\
z
W = – AeL
2
ede = -
1
AEL 2
(e 2 – e21 )
2
(3.22)
(f) Work Done in Changing the Area of a Surface Film
A film on the surface of a liquid has a surface tension, which is a property of the liquid and the
surroundings. The surface tension acts to make the surface area of the liquid a minimum. It has the unit of force per unit length. The work done on a homogeneous
liquid film in changing its surface area by an infinitesimal amount dA is
_
dW=–sdA
where s is the surface tension (N/m).
\
W=–
z
2
1
sdA
(3.23)
(g) Magnetization of a Paramagnetic Solid The work done per unit volume
on a magnetic material through which the magnetic and magnetization fields are
uniform is
_
d W = – HdI
and
W1–2 = –
z
I2
HdI
(3.24)
I1
where H is the field strength, and I is the component of the magnetization field in
the direction of the field. The minus sign provides that an increase in magnetization
(positive dI) involves negative work.
The following equations summarize the different forms of work transfer:
Displacement work
(compressible fluid)
W=
Electrical work
W=
z
z
z
2
pdV
1
2
E dC =
1
Shaft work
W=
Surface film
W=–
Stretched wire
W=–
EI dt
2
T dq
1
W=–
2
1
z
z
z
2
1
sdA
2
F dL
1
Magnetised solid
z
2
1
H dI
(3.25)
52
Engineering Thermodynamics
It may be noted in the above expressions that the work is equal to the integral of
the product of an intensive property and the change in its related extensive property.
These expressions are valid only for infinitesimally slow quasi-static processes.
There are some other forms of work which can be identified in processes that are
not quasi-static, for example, the work done by shearing forces in a process
involving friction in a viscous fluid.
3.5
FREE EXPANSION WITH ZERO WORK TRANSFER
Work transfer is identified only at the boundaries of a system. It is a boundary
phenomenon, and a form of energy in transit crossing the boundary. Let us consider
a gas separated from the vacuum by a partition (Fig. 3.17). Let the partition be
removed. The gas rushes to fill the entire volume. The expansion of a gas against
vacuum is called free expansion. If we neglect the work associated with the removal
of partition, and consider the gas and vacuum together as our system (Fig. 3.17a),
there is no work transfer involved here, since no work crosses the system boundary,
and hence
z
2 _
d W = 0,
z
although
1
2
1
pdV π 0
If only the gas is taken as the system (Fig. 3.17b), when the partition is removed
there is a change in the volume of the gas, and one is tempted to calculate the work
from the expression
z
2
pdV. However, this is not a quasi-static process, although
1
Partition
Gas
Partition
Boundary
Vacuum
Gas
Insulation
(a)
p
the initial and final end states are in equilibrium. Therefore, the work cannot be
calculated from this relation. The two end states can be located on the p–V diagram
and these are joined by a dotted line (Fig. 3.17c) to indicate that the process had
1
Vacuum
2
V
(b)
(c)
Partitions
p
1
2
Gas
V
Vacuum
(d)
(e)
Fig. 3.17 Free Expansion
53
Work and Heat Transfer
occurred. However, if the vacuum space is divided into a large number of small
volumes by partitions and the partitions are removed one by one slowly
(Fig. 3.17d), then every state passed through by the system is an equilibrium state
and the work done can then be estimated from the relation
z
2
pdV (Fig. 3.17e). Yet,
1
in free expansion of a gas, thee is no resistance to the fluid at the system boundary
as the volume of the gas increases to fill up the vacuum space. Work is done by a
system to overcome some resistance. Since vacuum does not offer any resistance,
there is no work transfer involved in free expansion.
3.6
NET WORK DONE BY A SYSTEM
Often different forms of work transfer occur simultaneously during a process
executed by a system. When all these work interactions have been evaluated, the
total or net work done by the system would be equal to the algebraic sum of these as
given below
Wtotal = Wdisplacement + Wshear + Welectrical + Wstirring + ...
3.7
HEAT TRANSFER
If a system has a non-adiabatic boundary, its temperature is not independent of the
temperature of the surroundings, and for the system between the states 1 and 2, the
_
work W depends on the path, and the differential d W is inexact. The work depends
on the terminal states 1 and 2 as well as the non-adiabatic path connecting them.
For consistency with the principle of conservation of energy, some other type of
energy transfer besides work must have taken place between the system and surroundings. This type of energy transfer must have occurred because of the temperature difference between the system and its surroundings, and it is identified as ‘heat’.
Thus, when an effect in a system occurs solely as a result of temperature difference
between the system and some other system, the process in which the effect occurs
shall be called a transfer of heat from the system at the higher temperature to the
system at the lower temperature.
Heat is defined as the form of energy that is transferred across a boundary by
virtue of a temperature difference. The temperature difference is the ‘potential’ or
‘force’ and heat transfer is the ‘flux’.
The transfer of heat between two bodies in direct contact is called conduction.
Heat may be transferred between two bodies separated by empty space or gases by
the mechanism of radiation through electromagnetic waves. A third method of heat
transfer is convection which refers to the transfer of heat between a wall and a fluid
system in motion.
The direction of heat transfer is taken from the high temperature system to the
low temperature system. Heat flow into a system is taken to be positive, and heat
54
Engineering Thermodynamics
flow out of a system is taken as negative
Boundary
(Fig. 3.18). The symbol Q is used for
heat transfer, i.e. the quantity of heat
Q
Surroundings
(Positive)
transferred within a certain time.
System
Heat is a form of energy in transit (like
Q
work transfer). It is a boundary
(Negative)
phenomenon, since it occurs only at the
boundary of a system. Energy transfer by
Fig. 3.18
Direction of Heat Transfer
virtue of temperature difference only is
called heat transfer. All other energy interactions may be termed as work transfer.
Heat is not that which inevitably causes a temperature rise. When heat is
transferred to an ice-and-water mixture, the temperature does not rise until all the
ice has melted. When a temperature rise in a system occurs, it may not be due to
heat transfer, since a temperature rise may be caused by work transfer also. Heat,
like work, is not a conserved quantity, and is not a property of a system.
A process in which no heat crosses the boundary of the system is called an
adiabatic process.
Thus, an adiabatic process is one in which there is only work interaction between
the system and its surroundings.
A wall which is impermeable to the flow of heat is an adiabatic wall, whereas a
wall which permits the flow of heat is a diathermic wall.
The unit of heat is Joule in S.I. units.
The rate of heat transfer or work transfer is given in kW or W.
3.8
HEAT TRANSFER—A PATH FUNCTION
The heat flow Q in a process can be quantified in terms of the work W done in the
same process between two given terminal states. Work W is different for different
non-adiabatic processes between these two states and it is not conserved. However,
from the principle of conservation of energy, the difference (Q – W) is conserved
for all paths between the same two states. The heat flow Q, like W, depends on the
process and is path-dependent and not a property.
Heat transfer is a path function, that is, the amount of heat transferred when a
system changes from state 1 to state 2 depends on the intermediate states through
which the system passes, i.e. its path. Therefore dQ is an inexact differential, and
we write
z
2_
1
d Q = Q1–2
or
1Q 2
The displacement work is given by
W1–2 =
z z
2_
2
1
1
dW =
pdV
55
Work and Heat Transfer
It is valid for a quasi-static process, and the work transfer involved is represented
by the area under the path on p-v diagram (Fig. 3.19a). Whenever there is a
difference in pressure, there will be displacement work. The pressure difference is
the cause and work transfer is the effect. The work transfer is equal to the integral
of the product of the intensive property, p and the differential change in the extensive
property, dV.
1
1
Quasi-Static
p
T
Quasi-Static
2
2
V
Area = Work transfer
X
Area = Heat transfer
2
W1 -2 =
Fig. 3.19
2
Q1 -2 =
pdV
Tdx
1
1
(a)
(b)
Representation of Work Transfer and Heat Transfer in Quasi-Static
Processes on p-v and T-x Coordinates
Likewise, whenever there is a difference in temperature, there will be heat flow.
The temperature difference is the cause and heat transfer is the effect. Just like
displacement work, the heat transfer can also be written as the integral of the product
of the intensive property T and the differential change of an extensive property, say
X (Fig. 3.19b).
Q1–2 =
z z
2_
2
1
1
dQ=
TdX
(3.26)
It must also be valid for a quasi-static process only, and the heat transfer involved
is represented by the area under the path 1–2 in T-X plot (Fig. 3.19b). Heat transfer
is, therefore, a path function, i.e. the amount of heat transferred when a system
changes from state 1 to state 2 depends on the path the system follows
_
(Fig. 3.19b). Therefore d Q is an inexact differential. Now,
_
d Q = T dX
where X is an extensive property and dX is an exact differential.
\
_
dX = 1 d Q
T
(3.27)
56
Engineering Thermodynamics
_
To make d Q integrable, i.e. an exact differential, it must be multiplied by an
integrating factor which is, in this case, 1/T. The extensive property X is yet to be
defined. It has been introduced in Chapter 7 and it is called ‘entropy’.
3.9
SPECIFIC HEAT AND LATENT HEAT
The specific heat of a substance is defined as the amount of heat required to raise a
unit mass of the substance through a unit rise in temperature. The symbol c will be
used for specific heat.
\
c=
Q J/kg K
m◊Dt
where Q is the amount of heat transfer (J), m, the mass of the substance (kg), and Dt,
the rise in temperature (K).
Since heat is not a property, as explained later, so the specific heat is qualified
with the process through which exchange of heat is made. For gases, if the process
is at constant pressure, it is cp, and if the process is at constant volume, it is cv. For
solids and liquids, however, the specific heat does not depend on the process. An
elegant manner of defining specific heats, cv and cp, in terms of properties is given
in Secs 4.5 and 4.6.
The product of mass and specific heat (mc) is called the heat capacity of the
substance. The capital letter C, Cp or Cv, is used for heat capacity.
The latent heat is the amount of heat transfer required to cause a phase change in
unit mass of a substance at a constant pressure and temperature. There are three
phases in which matter can exist: solid, liquid, and vapour or gas. The latent heat of
fusion (lfu ) is the amount of heat transferred to melt unit mass of solid into liquid, or
to freeze unit mass of liquid to solid. The latent heat of vaporization (lvap) is the
quantity of heat required to vaporize unit mass of liquid into vapour, or condense
unit mass of vapour into liquid. The latent heat of sublimation (lsub ) is the amount
of heat transferred to convert unit mass of solid to vapour or vice versa. lfu is not
much affected by pressure, whereas l vap is highly sensitive to pressure.
3 . 1 0 POINTS TO REMEMBER REGARDING HEAT
TRANSFER AND WORK TRANSFER
(a) Heat transfer and work transfer are the energy interactions. A closed system
and its surroundings can interact in two ways: by heat transfer and by work
transfer. Thermodynamics studies how these interactions bring about property changes in a system.
(b) The same effect in a closed system can be brought about either by heat transfer or by work transfer. Whether heat transfer or work transfer has taken place
depends on what constitutes the system.
(c) Both heat transfer and work transfer are boundary phenomena. Both are observed at the boundaries of the system, and both represent energy crossing the
boundaries of the system.
57
Work and Heat Transfer
(d) It is wrong to say ‘total heat’ or ‘heat content’ of a closed system, because heat
or work is not a property of the system. Heat, like work, cannot be stored by
the system. Both heat and work are the energy is transit.
(e) Heat transfer is the energy interaction due to temperature difference only. All
other energy interactions may be termed as work transfer.
(f) Both heat and work are path functions and inexact differentials. The magnitude of heat transfer or work transfer depends upon the path the system follows during the change of state.
Solved Examples
Example 3.1 An object of 40 kg mass falls freely under the influence of gravity
from an elevation of 100 m above the earth’s surface. The initial velocity is directed downward with a magnitude of 100 m/s. Ignoring the effect of air resistance,
what is the magnitude of the velocity, in m/s, of the object just before it strikes the
earth? The acceleration of gravity is g = 9.81 m/s2.
Solution Since the only force is that due to gravity,
m = 40 kg
1
1
mV22 + mgz2 = mV12 + mgZ1
2
2
V2 =
=
2gz1 + V12
V1 = 100 m/s
Z1 = 100 m
2 ¥ 9.81 ¥ 100 + 1002
g = 9.81 m/s2
Z2 = 0
= 104.4 m/s Ans.
The velocity increases, as expected, and the magnitude is independent of the
mass of the object.
Example 3.2 Gas from a bottle of compressed helium is used to inflate an
inelastic flexible balloon, originally folded completely flat to a volume of 0.5 m3. If
the barometer reads 760 mm Hg, what is the amount of work done upon the
atmosphere by the balloon? Sketch the system before and after the process.
Solution The firm line P1 (Fig. 3.20) shows the boundary of the system before
the process, and the dotted line P2 shows the boundary after the process. The
displacement work
Wd =
z
Balloon
pdV +
z
Bottle
pdV = p D V + 0
kN
¥ 0.5 m3 = 50.66 kJ
m2
This is positive, because work is done by the system. Work done by the
atmosphere is –50.66 kJ. Since the wall of the bottle is rigid, there is no pdV-work
involved in it.
= 101.325
58
Engineering Thermodynamics
P2
Final volume
of balloon = 0.5 m3
Valve
Balloon
initially flat
P2
P1
P = 760 mm Hg
= 101.325 kPa
Helium bottle
Fig. 3.20
It is assumed that the pressure in the balloon is atmospheric at all times, since the
balloon fabric is light, inelastic and unstressed. If the balloon were elastic and
stressed during the filling process, the work done by the gas would be greater than
50.66 kJ by an amount equal to the work done in stretching the balloon, although
the displacement work done by the atmosphere is still –50.66 kJ. However, if
the system includes both the gas and the balloon, the displacement work would be
50.66 kJ, as estimated above.
Example 3.3 When the valve of the evacuated bottle (Fig. 3.21) is opened,
atmospheric air rushes into it. If the atmospheric pressure is 101.325 kPa, and
0.6 m3 of air (measured at atmospheric conditions) enters into the bottle, calculate
the work done by air.
Valve
0.6 m3 of atm air
Initial boundary
Final boundary
Patm = 101.325 kPa
Fig. 3.21
59
Work and Heat Transfer
Solution
The displacement work done by air
Wd =
z
pdV +
Bottle
z
Free-air pdV
boundary
=0+pDV
= 101.325 kN/m2 ¥ 0.6 m3 = 60.8 kJ
Since the free-air boundary is contracting, the work done by the system is negative (DV being negative), and the surroundings do positive work upon the system.
Example 3.4 A piston and cylinder machine containing a fluid system has a
stirring device in the cylinder (Fig. 3.22). The piston is frictionless, and it is held
down against the fluid due to the atmospheric pressure of 101.325 kPa. The stirring device is turned 10,000 revolutions with an average torque against the fluid of
1.275 mN. Meanwhile the piston of 0.6 m diameter moves out 0.8 m. Find the net
work transfer for the system.
System
0.3 m
p = 101.325 kPa
W1
W2
W
Fig. 3.22
Solution
Work done by the stirring device upon the system (Fig. 3.22)
W1 = 2p TN
= 2p ¥ 1.275 ¥ 10,000 Nm = 80 kJ
This is negative work for the system.
Work done by the system upon the surroundings
W2 = (pA) ◊ L
kN p
¥ (0.6)2 m2 ¥ 0.80 m = 22.9 kJ
m2 4
This is positive work for the system. Hence, the net work transfer for the
system
W = W1 + W2 = – 80 + 22.9 = – 57.1 kJ
Example 3.5 The following data refer to a 12-cylinder, single-acting, two-stroke
marine diesel engine:
Speed—150 rpm
Cylinder diameter—0.8 m
= 101.325
60
Engineering Thermodynamics
Stroke of piston—1.2 m
Area of indicator diagram—5.5 ¥ 10–4 m2
Length of diagram—0.06 m
Spring value—147 MPa per m
Find the net rate of work transfer from the gas to the pistons in kW.
Solution
Mean effective pressure, pm, is given by
pm =
ad
¥ spring constant
ld
5.5 ¥ 10-4 m2
MPa
¥ 147
= 1.35 MPa
0.06 m
m
One engine cycle is completed in two strokes of the piston or one revolution of the
crank-shaft.
\ Work done in one minute
=
= pm LAN
= 1.35 ¥
p
(0.8)2 ¥ 1.2 ¥ 150 = 122 MJ
4
Since the engine is single-acting, and it has 12 cylinders, each contributing an equal
power, the rate of work transfer from the gas to the piston is given by
W = 122 ¥ 12 MJ/min
= 24.4 MJ/s
= 24.4 MW = 24,400 kW
Example 3.6 It is required to melt 5 tonnes/h of iron from a charge at 15°C to
molten metal at 1650°C. The melting point is 1535°C, and the latent heat is
270 kJ/kg. The specific heat in solid state is 0.502 and in liquid state (29.93/atomic
weight) kJ/kg K. If an electric furnace has 70% efficiency, find the kW rating
needed. If the density in molten state is 6900 kg/m3 and the bath volume is three
times the hourly melting rate, find the dimensions of the cylindrical
furnace if the length to diameter ratio is 2. The atomic weight of iron is 56.
Solution
Heat required to melt 1 kg of iron at 15°C to molten metal at 1650°C
= Heat required to raise the temperature from 15°C to 1535°C + Latent
heat + Heat required to raise the temperature from 1535°C to 1650°C
= 0.502 (1535 – 15) + 270 + 29.93 (1650 – 1535)/56
= 763 + 270 + 61.5 = 1094.5 kJ/kg
Melting rate = 5 ¥ 103 kg/h
So, the rate of heat supply required
= (5 ¥ 103 ¥ 1094.5) kJ/h
Work and Heat Transfer
61
Since the furnace has 70% efficiency, the rating of the furnace would be
=
Rate of heat supply per second
Furnace efficiency
=
5 ¥ 10 3 ¥ 1094.5
= 217 ¥ 103 kW
0.7 ¥ 3600
Volume needed =
3 ¥ 5 ¥ 10 3 3
m = 2.18 m3
6900
If d is the diameter and l the length of the furnace
p 2
d l = 2.18 m3
4
p 2
or
d ¥ 2d = 2.18 m3
4
\
d = 1.15 m
and
l = 2.30 m
Example 3.7 If it is desired to melt aluminium with solid state specific heat
0.9 kJ/kg K latent heat 390 kJ/kg, atomic weight 27, density in molten state
2400 kg/m3 and final temperature 700°C, find out how much metal can be
melted per hour with the above kW rating. Other data are as in the above
example. Also, find the mass of aluminium that the above furnace will hold. The
melting point of aluminium is 660°C.
Solution
Heat required per kg of aluminium
29.93
(700 – 660)
27
= 580.5 + 390 + 44.3 = 1014.8 kJ
= 0.9 (660 – 15) + 390 +
1014.8
= 1449.7 kJ/kg
0.7
With the given power, the rate at which aluminium can be melted
Heat to be supplied =
217 ¥ 10 3 ¥ 3600
kg/h = 5.39 tonnes/h
1449.7
Mass of aluminium that can be held in the above furnace
= 2.18 ¥ 2400 kg = 5.23 tonnes
=
Example 3.8 A cooling tower nozzle disperses water into a stream of droplets.
If the average diameter of the droplets is 60 microns, estimate the work required
for atomizing 1 kg of water isothermally at the ambient conditions. Given: surface
tension of water in contact with air = 0.07 N/m, density of water = 1000 kg/m3 .
Water is assumed to enter the nozzle through a pipe of 15-mm diameter.
Solution As water passes through the nozzle and comes out in the form of a spray
of droplets, there is a tremendous increase in the surface area. The process of at-
62
Engineering Thermodynamics
omization can be viewed as analogous to increase in the surface area of the film,
and the work done on the film is
f
W = Ú s dA = s (Af – Ai)
i
where Af = final area, Ai = initial area
If N = number of droplets with an average diameter of 60 mm,
N¥
\
N=
p
(60 ¥ 10–6)3 ¥ 1000 = 1 kg
6
6
p ¥ 216 ¥ 10-15 ¥ 1000
= 8.842 ¥ 109
Total surface area of all the droplets
Af = p (60 ¥ 10–6)2 ¥ 8.842 ¥ 109 = 100 m2
If we assume water enters the nozzle through a pipe of 15 mm diameter and 1 kg of
water is contained in L(m) length of the inlet pipe, then
p 2
d Lr = 1 kg
4
Surface area would be
S = pdL m2
4
p dL
S
=
=
= 0.267 m2/m
-3
p 2
15 ¥ 10 ¥ 1000
L
d Lr
4
Work done during atomization
W = (0.07) (100 – 0.287) = 6.98 J Ans.
Example 3.9 An electric motor drives a stirrer filted with a horizontal cylinder.
The cylinder of 40 cm diameter contains a fluid restrained by a frictionless piston.
During the stirring of fluid for 15 min the piston moves outward slowly by a distance of 30 cm against the atmospheric pressure of 1 bar. The current supplied to
the motor is 0.5 amp. from a 24-V lead-acid accumulator. If the conversion efficiency from electrical work to mechanical work output is 90%, estimate the work
done on the motor, stirrer and the atmosphere.
Dividing,
Solution
Work input to the motor
Wm = Ú V dQc = 24 ¥ 0.5 ¥ 15 ¥ 60
= 10800 J = 10.8 kJ Ans.
Work input to the stirrer
Ws = 10.8 ¥ 0.9 = 9.72 kJ Ans.
Work done by the fluid on the atmosphere
p
(0.4)2 ¥ 0.3 Nm
4
Ans.
W = p ◊ DV = 105 ¥
= 3.77 kJ
63
Work and Heat Transfer
The fluid receives a work input of 9.72 kJ from the stirrer and does 3.77 kJ of
work on the atmosphere at the moving foundary.
Example 3.10 A piston-cylinder device with air at an initial temperature of
30°C undergoes an expansion process for which pressure and volume are related
as given below:
p (kPa)
100
37.9
14.4
V (m3)
0.1
0.2
0.4
Calculate the work done by the system.
p1V1n = p2V2n = p3V3n
Solution
Here,
n=
ln (100 / 37.9)
ln ( p1 / p2 )
=
= 1.4
ln (0.2 / 0.1)
ln (V2 /V1)
Also,
n=
ln (37.9 /14.4)
ln ( p2 / p3 )
=
= 1.4
ln (0.4 / 0.2)
ln (V3 /V2 )
Therefore, n can be taken as 1.4 for the expansion process.
\
W=
p1V1 - p3V3
100 ¥ 0.1 - 14.4 ¥ 0.4
=
n -1
1.4 - 1
= 10.6 kJ
Ans.
Example 3.11 A piston-cylinder device operates 1 kg of fluid at 20 atm. pressure. The initial volume is 0.04 m3. The fluid is allowed to expand reversibly following a process pV1.45 = constant so that the volume becomes double. The fluid is
then cooled at constant pressure until the piston
1
comes back to the original position. Keeping the piston unaltered, heat is added reversibly to restore it to
pV 1.45 = C
the initial pressure. Calculate the work done in the
cycle.
p
Solution
2
3
p1 = 20 atm = 20 ¥ 1.01325 = 20.265 bar
V1 = 0.04 m 3, V2 = 2V1 = 0.08 m 3
V
p2 = (V1/V2)n ◊ p1 = (1/2)1.45 ¥ 20.265
Fig. 3.23
= 7.417 bar
W1–2 =
p1V1 - p2V2
100 [20.265 ¥ 0.04 - 7.417 ¥ 0.08]
=
n -1
0.45
= 47.82 kJ
W2–3 = p2 (V2 – V1) = – 100 ¥ 0.04 ¥ 7.417 = – 29.65 kJ
Work in the cycle = 47.82 – 29.65 = 18.17 kJ
Ans.
64
Engineering Thermodynamics
Summary
Work is done by a system if the sole effect on things external to the system could
be reduced to the raising of a weight. The weight may not actually be raised, but
the net effect external to the system would be the raising of a weight. Work done
by a system is taken to be positive, if it is done upon the system it is negative.
There are various types of work transfer crossing the boundary of a system:
(a) pdV work or displacement work, (b) electrical work, (c) shaft work, (d) stirring
or paddle-wheel work, (e) flow work, etc. In a stationary fluid system, the work
done by or upon the system is given by Ú pdV. If dV is positive, W is positive and
if dV is negative W is negative. An indicator diagram is similar to a pressurevolume plot, but it is a plot of pressure vs piston travel, and not a system pressurevolume diagram. The integration Ú pdV can only be done on a quasi-static path.
_
Work done is a path function and d W is an inexact differential.
Any transfer of energy to a system by virtue of temperature difference is known
as heat. Heat, like work, is a form of energy in transit and flows across a
boundary. Heat flow to a system is arbitrarily taken to be positive and heat flow
_
out of the system is regarded as negative. Heat is a path function and d Q is an
inexact differential.
The specific heat of a substance is defined as the amount of heat required to
raise the temperature of unit mass of the substance by 1°C. For a compressible
substance there can be two specific heats, cp and cv. The latent heat is the amount
of heat transfer required to cause a phase change at constant pressure of unit
mass of a substance like latent heats of fusion, vaporization and sublimation.
Review Questions
3.1 How can a closed system and its surroundings interact? What is the effect of
such interactions on the system?
3.2 When is work said to be done by a system?
3.3 What are positive and negative work interactions?
3.4 What is displacement work?
3.5 Under what conditions is the work done equal to
z
2
1
pdV?
3.6 What do you understand by path function and point function? What are exact
and inexact differentials?
3.7 Show that work is a path function, and not a property.
3.8 What is an indicator diagram?
3.9 What is mean effective pressure? How is it measured?
3.10 What are the indicated power and the brake power of an engine?
Work and Heat Transfer
65
3.11 How does the current flowing through a resistor represent work transfer?
3.12 What do you understand by flow work? Is it different from displacement work?
3.13 Why does free expansion have zero work transfer?
3.14 What is heat transfer? What are its positive and negative directions?
3.15 What are adiabatic and diathermic walls?
3.16 What is an integrating factor?
3.17 Show that heat is a path function and not a property.
3.18 What is the difference between work transfer and heat transfer?
3.19 Does heat transfer inevitably cause a temperature rise?
Problems
3.1 (a) A pump forces 1 m3/min of water horizontally from an open well to a closed
tank where the pressure is 0.9 MPa. Compute the work the pump must do
upon the water in an hour just to force the water into the tank against the
pressure. Sketch the system upon which the work is done before and after
the process.
Ans. 5400 kJ/h
(b) If the work done as above upon the water had been used solely to raise the
same amount of water vertically against gravity without change of pressure,
how many metres would the water have been elevated?
Ans. 91.74 m
(c) If the work done in (a) upon the water had been used solely to accelerate the
water from zero velocity without change of pressure or elevation, what velocity would the water have reached? If the work had been used to accelerate the water from an initial velocity of 10 m/s, what would the final velocity have been?
Ans. 42.4 m/s; 43.6 m/s
3.2 The piston of an oil engine, of area 0.0045 m2, moves downwards 75 mm, drawing in 0.00028 m3 of fresh air from the atmosphere. The pressure in the cylinder
is uniform during the process at 80 kPa, while the atmospheric pressure is
101.325 kPa, the difference being due to the flow resistance in the induction
pipe and the inlet valve. Estimate the displacement work done by the air finally
in the cylinder.
Ans. 27 J
3.3 An engine cylinder has a piston of area 0.12 m3 and contains gas at a pressure of
1.5 MPa. The gas expands according to a process which is represented by a
straight line on a pressure-volume diagram. The final pressure is 0.15 MPa. Calculate the work done by the gas on the piston if the stroke is 0.30 m.
Ans. 29.7 kJ.
3.4 A mass of 1.5 kg of air is compressed in a quasi-static process from 0.1 MPa to
0.7 MPa for which pv = constant. The initial density of air is 1.16 kg/m3. Find
the work done by the piston to compress the air.
Ans. 251.62 kJ
3.5 A mass of gas is compressed in a quasi-static process from 80 kPa, 0.1 m3 to
0.4 MPa, 0.03 m3. Assuming that the pressure and volume are related by
pvn = constant, find the work done by the gas system.
Ans. –11.83 kJ
66
Engineering Thermodynamics
3.6 A single-cylinder, double-acting, reciprocating water pump has an indicator diagram which is a rectangle 0.075 m long and 0.05 m high. The indicator spring
constant is 147 MPa per m. The pump runs at 50 rpm. The pump cylinder diameter is 0.15 m and the piston stroke is 0.20 m. Find the rate in kW at which the
piston does work on the water.
Ans. 43.3 kW
3.7 A single-cylinder, single-acting, 4 stroke engine of 0.15 m bore develops an indicated power of 4 kW when running at 216 rpm. Calculate the area of the indicator diagram that would be obtained with an indicator having a spring constant
of 25 ¥ 106 N/m3. The length of the indicator diagram is 0.1 times the length of
the stroke of the engine.
Ans. 505 mm2
3.8 A six-cylinder, 4-stroke gasoline engine is run at a speed of 2520 RPM. The area
of the indicator card of one cylinder is 2.45 ¥ 103 mm2 and its length is 58.5 mm.
The spring constant is 20 ¥ 106 N/m3. The bore of the cylinders is 140 mm and
the piston stroke is 150 mm. Determine the indicated power, assuming that each
cylinder contributes an equal power.
Ans. 243.57 kW
3.9 A closed cylinder of 0.25 m diameter is fitted with a light frictionless piston. The
piston is retained in position by a catch in the cylinder wall and the volume on
one side of the piston contains air at a pressure of 750 kN/m2. The volume on the
other side of the piston is evacuated. A helical spring is mounted coaxially with
the cylinder in this evacuated space to give a force of 120 N on the piston in this
position. The catch is released and the piston travels along the cylinder until it
comes to rest after a stroke of 1.2 m. The piston is then held in its position of
maximum travel by a ratchet mechanism. The spring force increases linearly with
the piston displacement to a final value of 5 kN. Calculate the work done by the
compressed air on the piston.
Ans. 3.07 kJ
3.10 A steam turbine drives a ship’s propeller through an 8 : 1 reduction gear. The
average resisting torque imposed by the water on the propeller is
750 ¥ 103 mN and the shaft power delivered by the turbine to the reduction gear
is 15 MW. The turbine speed is 1450 rpm. Determine (a) the torque developed by
the turbine, (b) the power delivered to the propeller shaft, and (c) the net rate of
working of the reduction gear.
Ans. (a) T = 98.84 km N, (b) 14.235 MW, (c) 0.765 MW
3.11 A fluid, contained in a horizontal cylinder fitted with a frictionless leakproof
piston, is continuously agitated by means of a stirrer passing through the cylinder cover. The cylinder diameter is 0.40 m. During the stirring process lasting 10
minutes, the piston slowly moves out a distance of 0.485 m against the
atmosphere. The net work done by the fluid during the process is 2 kJ. The speed
of the electric motor driving the stirrer is 840 rpm. Determine the torque in the
shaft and the power output of the motor.
Ans. 0.08 mN, 6.92 W
67
Work and Heat Transfer
p, bar
3.12 At the beginning of the compression stroke of a two-cylinder internal combustion engine the air is at a pressure of 101.325 kPa. Compression reduces the
volume to 1/5 of its original volume, and the law of compression is given by
pv1.2 = constant. If the bore and stroke of each cylinder is 0.15 m and 0.25 m,
respectively, determine the power absorbed in kW by compression strokes when
the engine speed is such that each cylinder undergoes 500 compression strokes
per minute.
Ans. 17.95 kW
3.13 Determine the total work done by a gas system following an expansion process as
A
B
shown in Fig. 3.23.
50
pv1-3 = C
Ans. 2.253 MJ
3.14 A system of volume V contains a mass m of
gas at pressure p and temperature T. The
C
macroscopic properties of the system obey
0.2 0.4
0.8
the following relationship:
FG p + a IJ (V – b) = mRT
H VK
2
V, m3
Fig.
3.23
where a, b, and R are constants.
Obtain an expression for the displacement work done by the system during a
constant-temperature expansion from volume V1 to volume V2. Calculate the
work done by a system which contains 10 kg of this gas expanding from 1 m3 to
10 m3 at a temperature of 293 K. Use the values a = 15.7 ¥ 10 Nm4, b = 1.07 ¥
10–2 m3, and R = 0.278 kJ/kg-K.
Ans. 1742 kJ
3.15 If a gas of volume 6000 cm3 and at pressure of 100 kPa is compressed
quasistatically according to pV2 = constant until the volume becomes 2000 cm3,
determine the final pressure and the work transfer.
Ans. 900 kPa, – 1.2 kJ
3.16 The flow energy of 0.124 m3/min of a fluid crossing a boundary to a system is
18 kW. Find the pressure at this point.
Ans. 8709 kPa
3.17 A milk chilling unit can remove heat from the milk at the rate of 41.87 MJ/h.
Heat leaks into the milk from the surroundings at an average rate of 4.187 MJ/h.
Find the time required for cooling a batch of 500 kg of milk from 45°C to 5°C.
Ans. 2h 13 min
Take the cp of milk to be 4.187 kJ/kg K.
3.18 680 kg of fish at 5°C are to be frozen and stored at – 12°C. The specific heat of
fish above freezing point is 3.182, and below freezing point is 1.717 kJ/kg K.
The freezing point is – 2°C, and the latent heat of fusion is 234.5 kJ/kg. How
much heat must be removed to cool the fish, and what per cent of this is latent
heat?
Ans. 186.28 MJ, 85.6%
3.19 A gas in a piston-cylinder assembly undergoes an expansion process for which
the relationship between pressure and volume is given by pV n = constant. The
initial pressure is 0.3 MPa, the initial volume is 0.1 M3 and the final volume is
0.2 m3. Determine the work for the process in kJ if (a) n = 1.5, (b) n = 1.0 and
(c) = n = 0.
Ans. (a) 17.6 kJ, (b) 20.79 kJ, (c) 30 kJ
68
Engineering Thermodynamics
3.20 A gas expands from an initial state with p1 = 340 kPa and V1 = 0.0425 m3 to a
final state where p2 = 136 kPa. If the pressure-volume relationship during the
process is pV2 = constant, determine the work in kJ.
Ans. 5.31 kJ
3.21 The drag force, Fd, imposed by the surrounding air on an automobile moving
with velocity V is given by
1
rV2
2
where Cd is a constant called the drag coefficient, A is the projected frontal area
of the vehicle, and r is the air density.
For Cd = 0.42, A = 2m2 and r = 1.23 kg/m3, calculate the power required in kW,
to overcome the drag at a constant velocity of 100 km/h.
Ans. 11.08 kW
3.22 An electric generator coupled to a windmill produces an average electrical power
output of 5 kW. The power is used to charge a storage battery. Heat transfer from
the battery to the surroundings occurs at a constant rate of 0.6 kW. Determine the
total amount of energy stored in the battery, in kJ, in 8h of operation.
Ans. 1.27 ¥ 105 kJ
Fd = Cd A
4
First Law of
Thermodynamics
4.1 FIRST LAW FOR A CLOSED SYSTEM UNDERGOING A
CYCLE
The transfer of heat and the performance of work may both cause the same effect
in a system. Heat and work are different forms of the same entity, called energy,
which is conserved. Energy which enters a system as heat may leave the system as
work, or energy which enters the system as work may leave as heat.
Let us consider a closed system which consists of a known mass of water
contained in an adiabatic vessel having a thermometer and a paddle wheel, as
shown in Fig. 4.1. Let a certain amount of work W1–2 be done upon the system by
the paddle wheel. The quantity of work can be measured by the fall of weight
which drives the paddle wheel through a pulley. The system was initially at
Pulley
Adiabatic
vessel
Weight
Fig. 4.1
Adiabatic Work
temperature t1, the same as that of atmosphere, and after work transfer let the
temperature rise to t2. The pressure is always 1 atm. The process 1–2 undergone by
the system is shown in Fig. 4.2 in generalized thermodynamic coordinates X, Y.
Let the insulation now be removed. The system and the surroundings interact by
70
Engineering Thermodynamics
y
heat transfer till the system returns to the original temperature t1, attaining the
condition of thermal equilibrium with the atmosphere. The amount of heat transfer
Q2–1 from the system during this process, 2–1, shown in Fig. 4.2, can be estimated.
The system thus executes a cycle,
which consists of a definite amount
2
of work input W1–2 to the system
followed by the transfer of an
amount of heat Q2–1 from the
W1-2
Q2-1
system. It has been found that this
1
W1–2 is always proportional to the
heat Q2–1, and the constant of
proportionality is called the Joule’s
x
equivalent or the mechanical
equivalent of heat. In the simple
Fig. 4.2
Cycle Completed by a System
with Two Energy Interactions:
example given here, there are only
Adiabatic Work Transfer W1–2
two energy transfer quantities as the
Followed by Heat Transfer Q2–1
system performs a thermodynamic
cycle. If the cycle involves many
more heat and work quantities, the same result will be found. Expressed
algebraically.
(S W)cycle = J (S Q)cycle
(4.1)
where J is the Joule’s equivalent. This is also expressed in the form
ÑÚ dW = J ÑÚ dQ
where the symbol ÑÚ denotes the cyclic integral for the closed path. This is the first
law for a closed system undergoing a cycle. It is accepted as a general law of
nature, since no violation of it has ever been demonstrated.
In the S.I. system of units, both heat and work are measured in the derived
unit of energy, the Joule. The constant of proportionality, J, is therefore unity
(J = 1 Nm/J).
The first law of thermodynamics owes much to J.P. Joule who, during the period
1840–1849, carried out a series of experiments to investigate the equivalence of
work and heat. In one of these experiments, Joule used an apparatus similar to the
one shown in Fig. 4.1. Work was transferred to the measured mass of water by
means of paddle wheel driven by the falling weight. The rise in the temperature of
water was recorded. Joule also used mercury as the fluid system, and later a solid
system of metal blocks which absorbed work by friction when rubbed against each
other. Other experiments involved the supplying of work in an electric current. In
every case, he found the same ratio (J) between the amount of work and the quantity
of heat that would produce identical effects in the system.
71
First Law of Thermodynamics
Prior to Joule, heat was considered to be an invisible fluid flowing from a body
of higher calorie to a body of lower calorie, and this was known as the caloric
theory of heat. It was Joule who first established that heat is a form of energy, and
thus laid the foundation of the first law of thermodynamics.
4.2 FIRST LAW FOR A CLOSED SYSTEM UNDERGOING A
CHANGE OF STATE
The expression (S W)cycle = (S Q)cycle applies only to systems undergoing cycles,
and the algebraic summation of all energy transfer across system boundaries is
zero. But if a system undergoes a change of state during which both heat transfer
and work transfer are involved, the net energy transfer will be stored or
accumulated within the system. If Q is the amount of heat transferred to the system
and W is the amount of work transferred from the system during the process
(Fig. 4.3), the net energy transfer (Q – W) will be stored in the system. Energy in
storage is neither heat nor work, and is given the name internal energy or simply,
the energy of the system.
Therefore
Q – W = DE
where DE is the increase in the energy of the system
or
Q = DE + W
(4.2)
Here Q, W, and DE are all expressed in the same units (in joules). Energy may be
stored by a system in different modes, as explained in Article 4.4.
If there are more energy transfer quantities involved in the process, as shown in
Fig. 4.4, the first law gives
(Q2 + Q3 – Q1) = DE + (W2 + W3 – W1 – W4)
Q3
W3
System
W
W1
Q
Surroundings
Fig. 4.3
Heat and Work
Interactions of a System
with Its Surroundings
in a Process
Fig. 4.4
Q2
System
Q1
W4
W2
Surroundings
System-surroundings
Interaction in a Process
Involving Many Energy
Fluxes
Energy is thus conserved in the operation. The first law is a particular
formulation of the principle of the conservation of energy. Equation (4.2) may also
be considered as the definition of energy. This definition does not give an absolute
value of energy E, but only the change of energy DE for the process. It can,
however, be shown that the energy has a definite value at every state of a system
and is, therefore, a property of the system.
72
ENERGY—A PROPERTY OF THE SYSTEM
Consider a system which changes its
state from state 1 to state 2 by following
the path A, and returns from state 2
to state 1 by following the path B
(Fig. 4.5). So the system undergoes a
cycle. Writing the first law for path A
QA = DEA + WA
1
p
4.3
Engineering Thermodynamics
B
C
A
(4.3)
2
and for path B
V
QB = DEB + WB
(4.4)
The processes A and B together
constitute a cycle, for which
Fig. 4.5 Energy—A Property
of a System
(S W)cycle = (S Q)cycle
or
WA + WB = QA + QB
or
QA – WA = WB – QB
(4.5)
From Eqs (4.3), (4.4), (4.5), it yields
DEA = – DEB
(4.6)
Similarly, had the system returned from state 2 to state 1 by following the path C
instead of path B
DEA = – DEC
(4.7)
From Eqs (4.6) and (4.7)
DEB = DEC
(4.8)
Therefore, it is seen that the change in energy between two states of a system is
the same, whatever path the system may follow in undergoing that change of state.
If some arbitrary value of energy is assigned to state 2, the value of energy at state
1 is fixed independent of the path the system follows. Therefore, energy has a
definite value for every state of the system. Hence, it is a point function and a
property of the system.
The energy E is an extensive property. The specific energy, e = E/M (J/kg), is an
intensive property.
The cyclic integral of any property is zero, because the final state is identical
with the initial state. ÑÚ dE = 0, ÑÚ dV = 0, etc. So for a cycle, the Eq. (4.2) reduces
to Eq. (4.1).
4.4
DIFFERENT FORMS OF STORED ENERGY
It was stated at the beginning that thermodynamics is the science of energy transfer
and its effect on the physical properties of a substance. Energy, as we know, is the
capacity of doing work. In thermodynamics, energy can be in two forms: (i) Energy
in transit, (ii) Energy in storage. Work and heat intractions are the forms of energy
First Law of Thermodynamics
73
in transit, observed at the boundaries of a system. They are not properties of a
system. They are path functions, their magnitudes depending upon the path the
system follows during a change of state. Energy in storage, called internal energy,
is a point or state function and hence a property of a system.
The symbol E refers to the total energy stored in a system. Basically there are
two modes in which energy may be stored in a system:
(a) Macroscopic energy mode
(b) Microscopic energy mode
The macroscopic energy mode includes the macroscopic kinetic energy and
potential energy of a system. Let us consider a fluid element of mass m having the
centre of mass velocity V (Fig. 4.6). The macroscopic kinetic energy EK of the
fluid element by virtue of its motion is given by
mV 2
2
If the elevation of the fluid element from an arbitrary datum is z, then the
macroscopic potential energy Ep by virtue of its position is given by
Ep = mgz
The microscopic energy mode refers to the energy stored in the molecular and
atomic structure of the system, which is called the molecular internal energy or
simply internal energy, customarily denoted by the symbol U. Matter is composed
of molecules. Molecules are in random thermal motion (for a gas) with an average
velocity v , constantly colliding with one another and with the walls (Fig. 4.6). Due
to a collision, the molecules may be subjected to rotation as well as vibration. They
can have translational kinetic energy, rotational kinetic energy, vibrational energy,
electronic energy, chemical energy and nuclear energy (Fig. 4.7). If e represents
the energy of one molecule, then
EK =
e = etrans + erot + evib + echem + eelectronic + enuclear
Random Thermal
Motion of Molecules
v
m
Fluid
Mass
Fig. 4.6
Z
Macroscopic and Microscopic Energy
(4.9)
74
Engineering Thermodynamics
v
Erot =
Etran =
1
/w 2
2
Evib =
1
kx 2
2
1
mV 2
2
Translational KE
Rotational KE
n
x
n
Vibrational Energy
Electron Spin
and Rotation
p
Nuclear binding energy
Fig. 4.7
Electron energy
Various Components of Internal Energy Stored in a Molecule
If N is the total number of molecules in the system, then the total internal energy
U = Ne
(4.10)
In an ideal gas there are no intermolecular forces of attraction and repulsion, and
the internal energy depends only on temperature. Thus
U = f (T) only
(4.11)
for an ideal gas.
Other forms of energy which can also be possessed by a system are magnetic
energy, electrical energy and surface (tension) energy. In the absence of these
forms, the total energy E of a system is given by
E = E K + EP + {
U
1
424
3
macro
(4.12)
micro
where EK, EP, and U refer to the kinetic, potential and internal energy, respectively.
In the absence of motion and gravity.
E K = 0, EP = 0
E=U
and Eq. (4.2) becomes
Q = DU + W
(4.13)
U is an extensive property of the system. The specific internal energy u is equal to
U/m and its unit is J/kg.
In the differential forms, Eqs (4.2) and (4.13) become
_
_
d Q = dE + d W
(4.14)
_
_
d Q = dU + d W
(4.15)
75
First Law of Thermodynamics
where
_
_
_
_
d W = d WpdV + d Wshaft + d Welectrical +
,
considering the different forms of work transfer which may be present. When only
pdV work is present, the equations become
_
d Q = dE + pdV
(4.16)
_
d Q = dU + pdV
(4.17)
or, in the integral form
4.5
Q = DE + Ú pdV
(4.18)
Q = DU + Ú pdV
(4.19)
SPECIFIC HEAT AT CONSTANT VOLUME
The specific heat of a substance at constant volume cV is defined as the rate of
change of specific internal energy with respect to temperature when the volume is
held constant, i.e.
Ê ∂u ˆ
cv = Á ˜
Ë ∂T ¯ v
(4.20)
For a constant-volume process
(D u)v = Ú
T2
T1
cv ◊ dT
(4.21)
The first law may be written for a closed stationary system composed of a unit
mass of a pure substance
Q = Du + W
_
_
or
d Q = du + d W
For a process in the absence of work other than pdV work
_
d W = pdV
_
\
(4.22)
d Q = dU + pdV
When the volume is held constant
(Q)v = (Du)v
\
(Q)v = Ú
T2
T1
cv dT
(4.23)
Heat transferred at constant volume increases the internal energy of the system.
If the specific heat of a substance is defined in terms of heat transfer, then
Ê ∂Q ˆ
cv = Á
Ë ∂T ˜¯ u
Since Q is not a property, this definition does not imply that cv is a property of a
substance. Therefore, this is not the appropriate method of defining the specific
_
heat, although ( d Q)v = du.
Since u, T, and v are properties, c v is a property of the system. The product
mcv = Cv is called the heat capacity at constant volume (J/K).
76
Engineering Thermodynamics
4.6 ENTHALPY
The enthalpy of a substance, h, is defined as
h = u + pv
(4.24)
It is an intensive property of a system (kJ/kg).
Internal energy change is equal to the heat transferred in a constant volume
process involving no work other than pdV work. From Eq. (4.22), it is possible to
drive an expression for the heat transfer in a constant pressure process involving
no work other than pdV work. In such a process in a closed stationary system of
unit mass of a pure substance
_
d Q = du + pdv
At constant pressure
\
or
or
pdv = d(pv)
_
( d Q)p = du + d(pv)
_
( d Q)p = d(u + pv)
_
( d Q)p = dh
(4.25)
where h = u + pv is the specific enthalpy, a property of the system.
Heat transferred at constant pressure increases the enthalpy of a system.
For an ideal gas, the enthalpy becomes
h = u + RT
(4.26)
Since the internal energy of an ideal gas depends only on the temperature
Eq. (4.11), the enthalpy of an ideal gas also depends on the temperature only, i.e.
h = f (T) only
(4.27)
Total enthalpy H = mh
Also
H = U + pV
and
h = H/m (J/kg)
4.7
SPECIFIC HEAT AT CONSTANT PRESSURE
The specific heat at constant pressure cp is defined as the rate of change of enthalpy
with respect to temperature when the pressure is held constant
cp =
FG ∂h IJ
H ∂T K
(4.28)
p
Since h, T and p are properties, so cp is a property of the system. Like cv, cp should
not be defined in terms of heat transfer at constant pressure, although
_
( d Q)p = dh.
For a constant pressure process
(Dh)p = Ú
T2
T1
cp dT
(4.29)
First Law of Thermodynamics
77
The first law for a closed stationary system of unit mass
_
d Q = du + pdv
Again
h = u + pv
\
dh = du + pdv + vdp
\
= dQ + vdp
_
d Q = dh – vdp
(4.30)
_
( d Q)p = dh
\
or
(Q)p = (Dh)p
\ From Eqs. (4.19) and (4.20)
T_
2
(Q)p =
cp dT
z
T1
cp is a property of the system, just like cv. The heat capacity at constant pressure Cp
is equal to mcp (J/K).
4.8 ENERGY OF AN ISOLATED SYSTEM
An isolated system is one in which there is no interaction of the system with the
_
_
surroundings. For an isolated system, d Q = 0, d W = 0.
The first law gives
dE = 0
or
E = constant
The energy of an isolated system is always constant.
4.9 PERPETUAL MOTION MACHINE OF THE
FIRST KIND—PMM1
The first law states the general principle of the conservation of energy. Energy is
neither created nor destroyed, but only gets transformed from one form to another.
There can be no machine which would continuously supply mechanical work
without some other form of energy disappearing simultaneously (Fig. 4.8). Such a
fictitious machine is called a perpetual motion machine of the first kind, or in brief,
PMM1. A PMM1 is thus impossible.
The converse of the above statement is also true, i.e. there can be no machine
which would continuously consume work without some other form of energy
appearing simultaneously (Fig. 4.9).
Q
Q
Engine
Fig. 4.8
A PMM1
W
Machine
W
Fig. 4.9 The Converse of PMM1
78
Engineering Thermodynamics
Solved Examples
Example 4.1 A stationary mass of gas is compressed without friction from an
initial state of 0.3 m3 and 0.105 MPa to a final state of 0.15 m3 and 0.105 MPa, the
pressure remaining constant during the process. There is a transfer of 37.6 kJ of
heat from the gas during the process. How much does the internal energy of the gas
change?
Solution
First law for a stationary system in a process gives
Q = DU + W
or
Q1–2 = U2 – U1 + W1–2
Here
W 1–2 = Ú
V2
V1
(1)
pdV = p(V2 – V1)
= 0.105 (0.15 – 0.30) MJ = – 15.75 kJ
Q1–2 = – 37.6 kJ
\ Substituting in Eq. (1)
– 37.6 kJ = U2 – U1 – 15.75 kJ
\
U2 – U1 = – 21.85 kJ
The internal energy of the gas decreases by 21.85 kJ in the process.
Example 4.2 When a system is taken from state a to state b, in Fig. 4.10 along
path acb, 84 kJ of heat flow into the system, and the system does 32 kJ of work. (a)
How much will the heat that flows into the system along path adb be, if the work
done is 10.5 kJ? (b) When the system is returned from b to a along the curved path,
the work done on the system is 21 kJ. Does the system absorb or liberate heat, and
how much of the heat is absorbed or liberated? (c) If Ua = 0 and Ud = 42 kJ, find the
heat absorbed in the processes ad and db.
Qacb = 84 kJ
p
Solution
c
b
a
d
Wacb = 32 kJ
We have
Qacb = Ub – Ua + Wacb
\
(a)
(b)
Ub – Ua = 84 – 32 = 52 kJ
Qadb = Ub – Ua + Wadb
= 52 + 10.5 = 62.5 kJ
Qb–a = Ua – Ub + Wb–a
= – 52 – 21 = – 73 kJ
The system liberates 73 kJ of heat
(c)
Wadb = Wad + Wdb = Wad = 10.5 kJ
V
Fig. 4.10
First Law of Thermodynamics
\
Qad = Ud – Ua + Wad
Now
= 42 – 0 + 10.5 = 52.5 kJ
Qadb = 62.5 kJ = Qad + Qdb
\
Qdb = 62.5 – 52.5 = 10 kJ
79
Example 4.3 A piston and cylinder machine contains a fluid system which
passes through a complete cycle of four processes. During a cycle, the sum of all
heat transfers is – 170 kJ. The system completes 100 cycles per min. Complete the
following table showing the method for each item, and compute the net rate of
work output in kW.
Process
a–b
b–c
c–d
d–a
Solution
Q (kJ/min)
0
21,000
–2,100
—
W (kJ/min)
2,170
0
—
—
Process a–b:
Q = DE + W
0 = DE + 2170
\
Process b–c:
DE = – 2170 kJ/min
Q = DE + W
21,000 = DE + 0
\
DE = 21,000 kJ/min
Process c–d:
Q = DE + W
– 2100 = – 36,600 + W
\
W = 34,500 kJ/min
Process d–a:
 Q = – 170 kJ
cycle
The system completes 100 cycles/min.
∵
Qab + Qbc + Qcd + Qda = – 17,000 kJ/min
\
0 + 21,000 – 2100 + Qda = – 17,000
Qda = – 35,900 kJ/min
Now ÑÚ dE = 0, since cyclic integral of any property is zero.
\
DEa – b + DEb – c + DEc – d + DEd – a = 0
– 2170 + 21,000 – 36,600 + DEd – a = 0
\
DEd – a = 17,770 kJ/min
DE (kJ/min)
—
—
–36,600
—
80
Engineering Thermodynamics
\
Wd – a = Qd – a – DEd – a
= – 35,900 – 17,770 = – 53,670 kJ/min
The table becomes
Process
a–b
b–c
c–d
d–a
Q (kJ/min)
0
21,000
–2100
–35,900
W (kJ/min)
2170
0
34,500
– 53,670
DE (kJ/min)
– 2170
21,000
– 36,600
17,770
 Q=  W
Since
cycle
cycle
Rate of work output
= – 17,000 kJ/min = – 283.3 kW
Example 4.4
equation
The internal energy of a certain substance is given by the following
u = 3.56 pv + 84
where u is given in kJ/kg, p is in kPa, and v is in m3/kg.
A system composed of 3 kg of this substance expands from an initial pressure of
500 kPa and a volume of 0.22 m3 to a final pressure 100 kPa in a process in which
pressure and volume are related by pv1.2 = constant.
(a) If the expansion is quasi-static, find Q, DU, and W for the process.
(b) In another process the same system expands according to the same pressurevolume relationship as in part (a), and from the same initial state to the same
final state as in part (a), but the heat transfer in this case is 30 kJ. Find the
work transfer for this process.
(c) Explain the difference in work transfer in parts (a) and (b).
Solution
(a)
u = 3.56 pv + 84
Du = u2 – u1 = 3.56 (p2v2 – p1v1 )
\
Now
\
DU = 3.56 (p2V2 – p1V1 )
p 1V 11.2 = p2V21.2
Êp ˆ
V 2 = V1 Á 1 ˜
Ëp ¯
2
\
1/1.2
Ê 5ˆ
= 0.22 Á ˜
Ë 1¯
1/1.2
= 0.22 ¥ 3.83 = 0.845 m3
DU = 356 (1 ¥ 0.845 – 5 ¥ 0.22) kJ
= – 356 ¥ 0.255 = – 91 kJ
For a quasi-static process
W = Ú pdV =
p2 V2 – p1V1
1– n
First Law of Thermodynamics
=
81
a1 ¥ 0.845 – 5 ¥ 0.22f 100
1 – 1.2
= 127.5 kJ
\
Q = DU + W
= – 91 + 127.5 = 36.5 kJ
(b) Here Q = 30 kJ
Since the end states are the same, DU would remain the same as in (a).
\
W = Q – DU
= 30 – (– 91) = 121 kJ
(c) The work in (b) is not equal to Ú pdV since the process is not quasi-static.
Example 4.5 A fluid is confined in a cylinder by a spring-loaded, frictionless
piston so that the pressure in the fluid is a linear function of the volume
( p = a + bV ). The internal energy of the fluid is given by the following equation
U = 34 + 3.15 pV
where U is in kJ, p in kPa, and V in cubic metre. If the fluid changes from an initial
state of 170 kPa, 0.03 m3 to a final state of 400 kPa, 0.06 m3, with no work other
than that done on the piston, find the direction and magnitude of the work and heat
transfer.
Solution The change in the internal energy of the fluid during the process.
U2 – U1 = 3.15 ( p2V2 – p1V1)
= 315 (4 ¥ 0.06 – 1.7 ¥ 0.03)
= 315 ¥ 0.189 = 59.5 kJ
Now
p = a + bV
170 = a + b ¥ 0.03
400 = a + b ¥ 0.06
From these two equations
a = – 60 kN/m2
b = 7667 kN/m5
Work transfer involved during the process
W 1–2 = Ú
V2
V1
V2
pdV = Ú (a + bV ) dV
V1
= a(V2 – V1) + b
V22 - V12
2
82
Engineering Thermodynamics
b
È
˘
= (V2 – V1) Í a + (V1 + V2 ) ˙
2
Î
˚
= 0.03 m3
LM - 60 kN/m + 7667 kN ¥ 0.09 m OP
2 m
N
Q
2
3
5
= 8.55 kJ
Work is done by the system, the magnitude being 8.55 kJ.
\ Heat transfer involved is given by
Q1–2 = U2 – U1 + W1–2
= 59.5 + 8.55 = 68.05 kJ
68.05 kJ of heat flow into the system during the process.
p
Example 4.6 A stationary fluid system goes through a cycle shown in Fig. 4.11
comprising the following processes:
(i) Process 1–2 isochoric heat addition of 235 kJ/kg;
(ii) Process 2–3 adiabatic expansion to its original pressure with loss of
70 kJ/kg in internal energy;
(iii) Process 3–1 isobaric compression to its original volume with heat rejection
of 200 kJ/kg.
Prepare a balance sheet of energy
quantities and find the overall changes
2
during the cycle.
Solution
Q1–2 = 235 kJ/kg, W1–2 = 0, u2 – u1
= Q1–2 – W1– 2 = 235 kJ/kg
1
3
Q2–3 = 0, u3 – u2 = – 70 kJ/kg; W2– 3
= Q2– 3 – (u3 –u2) = 0 – (– 70)
v
Fig. 4.11
= 70 kJ/kg
Q3–1 = – 200 kJ/kg, u1 – u3 = (u1 – u2) – (u3 – u2) = – 235 + 70
= – 165 kJ/kg
W 3–1 = Q3– 1 – (u1 – u3 ) = – 200 – (– 165) = – 35 kJ/kg
 Q = Q1- 2 + Q2–3 + Q3– 1 = 235 + 0 – 200 = 35 kJ/kg
cycle
 W = W1–2 + W2– 3 + W3–1 = 0 + 70 – 35 = 35 kJ/kg
cycle
\
ÂQ = ÂW
cycle
cycle
This cycle is known as Lenoir cycle, and the operation of pulse jet approximates
to it.
First Law of Thermodynamics
83
Summary
The first law of thermodynamics is a special statement of the law of conservation
of energy. It may be written for a cycle
(SQ)cycle = (SW)cycle
or for a process.
Q = DE + W
= DEK + DEp + DU + W
1
mV2 + mgZ + DU + W
2
where E is the amount of energy stored in a system and it is a property of the
system. The stored energy can be either in macroscopic mode like kinetic and
potential energy (EK and EP) and also in microscopic mode called intergy energy
(U) where energy is stored in the molecules of a system.
=
Ê ∂u ˆ
. The heat supplied
Ë ∂T ˜¯ v
The specific heat at constant volume is defined as cv = Á
to a system at constant volume increases its internal energy. The enthalpy of a
substance h is defined as h = u + pv. The specific heat at constant pressure is
Ê ∂h ˆ
. The heat supplied to a system at constant pressure
Ë ∂T ˜¯ p
defined as cp = Á
increases its enthalpy. The heat capacities are equal to Cv = mcv and Cp = mcp.
The energy of an isolated system is always constant.
Review Questions
4.1 State the first law for a closed system undergoing a cycle.
4.2 What was the contribution of J.P. Joule in establishing the first law?
4.3 What is the caloric theory of heat? Why was it rejected?
4.4 Which is the property introduced by the first law?
4.5 State the first law for a closed system undergoing a change of state.
4.6 Show that energy is a property of a system.
4.7 What are the modes in which energy is stored in a system?
4.8 Define internal energy. How is energy stored in molecules and atoms?
4.9 What is the difference between the standard symbols of E and U?
4.10 What is the difference between heat and internal energy?
4.11 Define enthalpy. Why does the enthalpy of an ideal gas depend only on temperature?
4.12 Define the specific heats at constant volume and at constant pressure.
4.13 Why should specific heat not be defined in terms of heat transfer?
84
Engineering Thermodynamics
4.14 Which property of a system increases when heat is transferred: (a) at constant
volume, (b) at constant pressure?
4.15 What is a PMM1? Why is it impossible?
Problems
4.1 An engine is tested by means of a water brake at 1000 rpm. The measured torque
of the engine is 10000 mN and the water consumption of the brake is 0.5 m3/s, its
inlet temperature being 20°C. Calculate the water temperature at exit, assuming
that the whole of the engine power is ultimately transformed into heat which is
absorbed by the cooling water.
Ans. 20.5°C
4.2 In a cyclic process, heat transfers are + 14.7 kJ, – 25.2 kJ, – 3.56 kJ and + 31.5
kJ. What is the net work for this cyclic process?
Ans. 17.34 kJ
4.3 A slow chemical reaction takes place in a fluid at the constant pressure of
0.1 MPa. The fluid is surrounded by a perfect heat insulator during the reaction
which begins at state 1 and ends at state 2. The insulation is then removed and
105 kJ of heat flow to the surroundings as the fluid goes to state 3. The following
data are observed for the fluid at states 1, 2 and 3.
t(°C )
State
V(m3)
1
0.003
20
2
0.3
370
3
0.06
20
For the fluid system, calculate E2 and E3, if E1 = 0
Ans. E2 = – 29.7 kJ, E3 = – 110.7 kJ
4.4 During one cycle the working fluid in an engine engages in two work interactions: 15 kJ to the fluid and 44 kJ from the fluid, and three heat interactions, two
of which are known: 75 kJ to the fluid and 40 kJ from the fluid. Evaluate the
magnitude and direction of the third heat transfer.
Ans. – 6 kJ
4.5 A domestic refrigerator is loaded with food and the door closed. During a certain
period the machine consumes 1 kW h of energy and the internal energy of the
system drops by 5000 kJ. Find the net heat transfer for the system.
Ans. – 8.6 MJ
4.6 1.5 kg of liquid having a constant specific heat of 2.5 kJ/kg K is stirred in a wellinsulated chamber causing the temperature to rise by 15°C. Find DE and W for
the process.
Ans. DE = 56.25 kJ, W = – 56.25 kJ
4.7 The same liquid as in Problem 4.6 is stirred in a conducting chamber. During the
process 1.7 kJ of heat are transferred from the liquid to the surroundings, while
the temperature of the liquid is rising to 15°C. Find DE and W for the process.
Ans. DE = 54.55 kJ, W = 56.25 kJ
4.8 The properties of a certain fluid are related as follows
u = 196 + 0.718 t
pv = 0.287 (t + 273)
where u is the specific internal energy (kJ/kg), t is in °C, p is pressure (kN/m2),
and v is specific volume (m3/kg). For this fluid, find cv and cp.
Ans. 0.718, 1.005 kJ/kg K
First Law of Thermodynamics
85
4.9 A system composed of 2 kg of the above fluid expands in a frictionless piston
and cylinder machine from an initial state of 1 MPa, 100°C to a final temperature of 30°C. If there is no heat transfer, find the net work for the process.
Ans. 100.52 kJ
4.10 If all the work in the expansion of Problem 4.9 is done on the moving piston,
show that the equation representing the path of the expansion in the pv-plane is
given by pv1.4 = constant.
4.11 A stationary system consisting of 2 kg of the fluid of Problem 4.8 expands in an
adiabatic process according to pv 1.2 = constant. The initial conditions are 1 MPa
and 200°C, and the final pressure is 0.1 MPa. Find W and DE for the process.
Why is the work transfer not equal to Ú pdV?
Ans. W = 217.35, DE = – 217.35 kJ, Ú pdV = 434.4 kJ
4.12 A mixture of gases expands at constant pressure from 1 MPa, 0.03 m3 to
0.06 m3 with 84 kJ positive heat transfer. There is no work other than that
done on a piston. Find DE for the gaseous mixture.
Ans. 54 kJ
The same mixture expands through the same state path while a stirring device
does 21 kJ of work on the system. Find DE, W, and Q for the process.
Ans. 54 kJ, – 21 kJ, 33 kJ
4.13 A mass of 8 kg gas expands within a flexible container so that the p–v relationship is of the form pv1.2 = const. The initial pressure is 1000 kPa and the
initial volume is 1 m3. The final pressure is 5 kPa. If specific internal energy
of the gas decreases by 40 kJ/kg, find the heat transfer in magnitude and
direction.
Ans. + 2615 kJ
4.14 A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p = a + bV, where a and b are constants. The initial and final pressures
are 1000 kPa and 200 kPa respectively and the corresponding volumes are
0.20 m3 and 1.20 m3. The specific internal energy of the gas is given by the relation
u = 1.5 pv – 85 kJ/kg
where p is the kPa and v is in m3/kg. Calculate the net heat transfer and the
maximum internal energy of the gas attained during expansion.
Ans. 660 kJ, 503.3 kJ
4.15 The heat capacity at constant pressure of a certain system is a function of temperature only and may be expressed as
Cp = 2.093 +
41.87
J/°C
t + 100
where t is the temperature of the system in °C. The system is heated while it is
maintained at a pressure of 1 atmosphere until its volume increases from
2000 cm3 to 2400 cm3 and its temperature increases from 0°C to 100°C. (a) Find
the magnitude of the heat interaction. (b) How much does the internal energy of
the system increase?
Ans. (a) 238.32 J (b) 197.79 J
4.16 An imaginary engine receives heat and does work on a slowly moving piston at
such rates that the cycle of operation of 1 kg of working fluid can be represented
as a circle 10 cm in diameter on a p–v diagram on which 1 cm = 300 kPa and
86
Engineering Thermodynamics
1 cm = 0.1 m3/kg. (a) How much work is done by each kg of working fluid for
each cycle of operation? (b) The thermal efficiency of an engine is defined as the
ratio of work done and heat input in a cycle. If the heat rejected by the engine in
a cycle is 1000 kJ per kg of working fluid, what would be its thermal efficiency?
Ans. (a) 2356.19 kJ/kg, (b) 0.702
4.17 A gas undergoes a thermodynamic cycle consisting of three processes
beginning at an initial state where p1 = 1 bar, V1 = 1.5 m3 and U1 = 512 kJ. The
processes are as follows:
(i) Process 1–2: Compression with pV = constant to p2 = 2 bar, U2 = 690 kJ
(ii) Process 2–3: W23 = 0, Q23 = – 150 kJ, and
(iii) Process 3–1: W31 = + 50 kJ. Neglecting KE and PE changes, determine the
heat interactions Q12 and Q31.
Ans. 74 kJ, 22 kJ
4.18 A gas undergoes a thermodynamic cycle consisting of the following
processes: (i) Process 1–2: Constant pressure p = 1.4 bar, V1 = 0.028 m3, W12 =
10.5 kJ, (ii) Process 2–3: Compression with pV = constant, U3 = U2, (iii) Process
3–1: Constant volume, U1 – U3 = – 26.4 kJ. There are no significant changes in
KE and PE. (a) Sketch the cycle on a p-V diagram. (b) Calculate the net work for
the cycle in kJ. (c) Calculate the heat transfer for process 1–2 (d) Show that
S Q = SW.
cycle
cycle
Ans. (b) – 8.28 kJ, (c) 36.9 kJ
4.19 A fluid contained in a cylinder receives 150 kJ of mechanical energy by means
of a paddle wheel, together with 50 kJ in the form of heat. At the same time, a
piston in the cylinder moves in such a way that the pressure remains constant at
200 kN/m2 during the fluid expansion from 2m3 to 5m3. What is the change
in internal energy, and in enthalpy?
Ans. – 400 kJ, + 200 kJ
5
First Law Applied to
Flow Processes
5.1 CONTROL VOLUME
For any system and in any process, the first law can be written as
Q = DE + W
where E represents all forms of energy stored in the system.
For a pure substance,
E = EK + EP + U
where EK is the K.E. EP the P.E. and U the residual energy stored in the molecular
structure of the substance.
\
Q = DEK + DEP + DU + W
(5.1)
When there is mass transfer across the system boundary, the system is called an
open system. Most of the engineering devices are open systems involving the flow
of fluids through them.
Equation (5.1) refers to a system having a particular mass of substance, and it is
free to move from place to place.
Consider a steam turbine (Fig. 5.1) in which steam enters at a high pressure, does
work upon the turbine rotor, and then
leaves the turbine at low pressure
Moving system
through the exhaust pipe.
Control
If a certain mass of steam is consurface
sidered as the thermodynamic system,
W
then the energy equation becomes
Control
Q = DEK + DEP + DU + W
and in order to analyze the expansion
process in turbine the moving system
is to be followed as it travels through
the turbine, taking into account the
work and heat interactions all the way
through. This method of analysis is
similar to that of Lagrange in fluid
mechanics.
volume
Turbine
Shaft
Q
Fig. 5.1
Exhaust pipe
Flow Process Involving Work and
Heat Interactions
88
Engineering Thermodynamics
Although the system approach is quite valid, there is another approach which is
found to be highly convenient. Instead of concentrating attention upon a certain
quantity of fluid, which constitutes a moving system in flow processes, attention is
focused upon a certain fixed region in space called a control volume through which
the moving substance flows. This is similar to the analysis of Euler in fluid mechanics.
To distinguish the two concepts, it may be noted that while the system (closed)
boundary usually changes shape, position and orientation relative to the observer,
the control volume boundary remains fixed and unaltered. Again, while matter usually crosses the control volume boundary, no such flow occurs across the
system boundary.
The broken line in Fig. 5.1 represents the surface of the control volume which is
known as the control surface. This is the same as the system boundary of the open
system. The method of analysis is to inspect the control surface and account for all
energy quantities transferred through this surface. Since there is mass transfer
across the control surface, a mass balance also has to be made. Sections 1 and 2
allow mass transfer to take place, and Q and W are the heat and work interactions
respectively.
5.2 STEADY FLOW PROCESS
As a fluid flows through a certain control volume, its thermodynamic properties may
vary along the space coordinates as well as with time. If the rates of flow of mass and
energy through the control surface change with time, the mass and energy within
the control volume also would change with time.
‘Steady flow’ means that the rates of flow of mass and energy across the control
surface are constant.
In most engineering devices, there is a constant rate of flow of mass and energy
through the control surface, and the control volume in course of time attains a
steady state. At the steady state of a system, any thermodynamic property will have
a fixed value at a particular location, and will not alter with time. Thermodynamic
properties may vary along space coordinates, but do not vary with time. ‘Steady
state’ means that the state is steady or invariant with time.
5.3 MASS BALANCE AND ENERGY BALANCE IN A SIMPLE
STEADY FLOW PROCESS
In Fig. 5.2, a steady flow system has been shown in which, one stream of fluid enters
and an another stream leaves the control volume. There is no accumulation of mass
or energy within the control volume, and the properties at any location within the
control volume are steady with time. Sections 1.1 and 2.2 indicate, respectively, the
entrance and exit of the fluid across the control surface. The following quantities are
defined with reference to Fig. 5.2
A1, A2—cross-section of stream, m 2
w1, w2—mass flow rate, kg/s
p1, p2—pressure, absolute, N/m2
89
First Law Applied to Flow Processes
dQ
dt
Flow out
w2
Shaft
Steady flow device
dWx
dt
Flow in
w1
Z1
Control volume
Z2
C.S.
Datum
Fig. 5.2 Steady Flow Process
v1, v2—specific volume, m 3/kg
u1, u2—specific internal energy, J/kg
V1, V2—velocity, m/s
Z1, Z2—elevation above an arbitrary datum, m
_
dQ
—net rate of heat transfer through the control surface, J/s
dt
_
d Wx
—net rate of work transfer through the control surface, J/s
dt
exclusive of work done at Sections 1 and 2 in transferring the fluid through the
control surface.
t—time, s.
Subscripts 1 and 2 refer to the inlet and exit sections.
5.3.1
Mass Balance
By the conservation of mass, if there is no accumulation of mass within the control
volume, the mass flow rate entering must equal the mass flow rate leaving, or
w1 = w2
A1V1 A2 V2
=
v2
v1
This equation is known as the equation of continuity.
or
5.3.2
(5.2)
Energy Balance
In a flow process, the work transfer may be of two types: the external work and the
flow work. The external work refers to all the work transfer across the control surface other than that due to normal fluid forces. In engineering thermodynamics the
only kinds of external work of importance are shear (shaft or stirring) work and
electrical work. In Fig. 5.2 the only external work occurs in the form of shaft work,
Wx. The flow work, as discussed in Sec. 3.4, is the displacement work done by the
90
Engineering Thermodynamics
fluid of mass dm1 at the inlet Section 1 and that of mass dm2 at the exit Section 2,
which are (– p1v1 dm1) and (+ p2v2 dm2) respectively. Therefore, the total work
transfer is given by
W = Wx – p1v1dm1 + p2v2dm2
(5.3)
In the rate form,
_
_
dW
d Wx
dm1
dm2
=
– p1 v 1
+ p 2 v2
dt
dt
dt
dt
_
_
dW
d Wx
or
=
– w1p1v1 + w2p2v2
(5.4)
dt
dt
Since there is no accumulation of energy, by the conservation of energy, the total
rate of flow of all energy streams entering the control volume must equal the total
rate of flow of all energy streams leaving the control volume. This may be expressed
in the following equation
_
_
dQ
dW
w1e1 +
= w2e2 +
dt
dt
_
dW
Substituting for
from Eq. (5.4)
dt
_
_
dQ
d Wx
w1e1 +
= w2e2 +
– w1 p1 v 1 + w 2 p 2 v 2
dt
dt
_
_
dQ
d Wx
\
w1e1 + w1p1v1 +
= w2e2 + w2p2v2 +
(5.5)
dt
dt
where e1 and e2 refer to the energy carried into or out of the control volume with
unit mass of fluid.
The specific energy e is given by
e = ek + ep + u =
V2
+ Zg + u
2
(5.6)
Substituting the expression for e in Equation (5.5)
FG V + Z g + u IJ + w p v + d_ Q
dt
H2
K
F V + Z g + u IJ + w p v + d_ W
=w G
dt
H2
K
F V + Z gIJ + d_ Q
w Gh +
H 2 K dt
F V + Z gIJ + d_ W
= w Gh +
H 2 K dt
w1
2
or
2
1
1
1
1 1 1
2
2
2 2 2
2
2
x
2
1
1
1
2
2
1
2
2
x
2
where h = u + pv.
And, since
w1 = w2, let w = w1 = w2 =
dm
dt
(5.7)
91
First Law Applied to Flow Processes
Dividing Equation (5.7) by
dm
dt
_
_
dQ
d Wx
V12
V2
+ Z1g +
= h2 + 2 + Z 2g +
(5.8)
dm
dm
2
2
Equations (5.7) and (5.8) are known as steady flow energy equations (S.F.E.E.), for a
single stream of fluid entering and a single stream of fluid leaving the control volume. All the terms in Equation (5.8) represent energy flow per unit mass of fluid
(J/kg), whereas the terms in Equation (5.7) represent energy flow per unit time (J/s).
The basis of energy flow per unit mass is usually more convenient when only a
single stream of fluid enters and leaves a control volume. When more than one fluid
stream is involved the basis of energy flow per unit time is more convenient.
Equation (5.8) can be written in the following form,
h1 +
V22 - V12
+ g(Z2 – Z1)
(5.9)
2
where Q and Wx refer to energy transfer per unit mass. In the differential form, the
SFEE becomes
_
_
d Q – d Wx = dh + V d V + gdZ
(5.10)
When more than one stream of fluid enters or leaves the control volume
(Fig. 5.3), the mass balance and energy balance for steady flow are given as
follows.
Q – Wx = (h2 – h1) +
3
w3
w1
3
C.V.
dWx
dt
4
w2
w4
4
dQ
dt
Fig. 5.3
C.S.
Steady Flow Process Involving Two Fluid Streams at the Inlet
and Exit of the Control Volume
Mass balance
w1 + w2 = w3 + w4
(5.11)
A1V1 A2 V2
AV
AV
+
= 3 3+ 4 4
v1
v2
v3
v4
Energy balance
(5.12)
F V + Z gI + w F h + V + Z gI + d_ Q
GH 2 JK GH 2
JK dt
_
F
I
F
I
dW
V
V
= w Gh +
H 2 + Z gJK + w GH h + 2 + Z gJK + dt
w1 h1 +
2
1
2
2
1
2
2
3
4
4
2
3
3
3
2
2
4
x
4
(5.13)
92
Engineering Thermodynamics
The steady flow energy equation applies to a wide variety of processes like pipe
line flows, heat transfer processes, mechanical power generation in engines and
turbines, combustion processes, and flows through nozzles and diffusors. In certain problems, some of the terms in steady flow energy equation may be negligible
or zero. But it is best to write the full equation first, and then eliminate the terms
which are unnecessary.
5.4
SOME EXAMPLES OF STEADY FLOW PROCESSES
The following examples illustrate the applications of the steady flow energy equation in some of the engineering systems.
5.4.1
Nozzle and Diffusor
A nozzle is a device which increases the velocity or K.E. of a fluid at the expense of
its pressure drop, whereas a diffusor increases the pressure of a fluid at the expense
of its K.E. Figure 5.4 shows a nozzle which is insulated. The steady flow energy
equation of the control surface gives
_
_
d Wx
dQ
V2
V2
h1 + 1 + Z1 g +
= h2 + 2 + Z2g +
dm
dm
2
2
2
1
C.S.
m
m
1
Insulation
2
Fig. 5.4
Steady Flow Process Involving Two Fluid Streams at the Inlet and Exit of
the Control Volume
_
_
d Wx
dQ
Here
= 0,
= 0, and the change in potential energy is zero. The equadm
dm
tion reduces to
V2
V2
h1 + 1 = h2 + 2
(5.14)
2
2
The continuity equation gives
AV
AV
w= 1 1= 2 2
(5.15)
v1
v2
When the inlet velocity or the ‘velocity of approach’ V1 is small compared to the exit
velocity V2, Eq. (5.14) becomes
h1 = h2 +
V22
2
First Law Applied to Flow Processes
or
V2 =
93
2 ( h1 - h2 ) m/s
where (h1 – h2) is in J/kg.
Equations (5.14) and (5.15) hold good for a diffusor as well.
5.4.2
Throttling Device
When a fluid flows through a constricted passage, like a partially opened valve,
an orifice, or a porous plug, there is an appreciable drop in pressure, and the flow
is said to be throttled. Figure 5.5 shows the process of throttling by a partially
opened valve on a fluid flowing in an insulated pipe. In the steady-flow energy
Equation (5.8),
_
_
d Wx
dQ
= 0,
=0
dm
dm
and the changes in P.E. are very small and ignored. Thus, the S.F.E.E. reduces to
h1 +
V12
V2
= h2 + 2
2
2
1
1
Insulation
Control surface
Fig. 5.5
Flow Through a Valve
Often the pipe velocities in throttling are so low that the K.E. terms are also
negligible. So
h1 = h2
(5.16)
or the enthalpy of the fluid before throttling is equal to the enthalpy of the fluid
after throttling.
5.4.3
Turbine and Compressor
Turbines and engines give positive power output, whereas compressors and pumps
require power input.
For a turbine (Fig. 5.6) which is well insulated, the flow velocities are often small,
and the K.E. terms can be neglected. The S.F.E.E. then becomes
_
d Wx
h 1 = h2 +
dm
Wx
or
= (h1 – h2)
m
94
Engineering Thermodynamics
It is seen that work is done by the
fluid at the expense of its enthalpy.
Similarly, for an adiabatic pump or
compressor, work is done upon the
fluid and W is negative. So the S.F.E.E.
becomes
W
h 1 = h2 – x
m
Wx
or
= h2 – h1
m
The enthalpy of the fluid increases
by the amount of work input.
5.4.4
m
C.S.
1
WT
Turbine
Insulation
m
Fig. 5.6
Flow Through a Turbine
Heat Exchanger
A heat exchanger is a device in which heat is transferred from one fluid to another.
Figure 5.7 shows a steam condenser, where steam condensers outside the tubes and
cooling water flows through the tubes. The S.F.E.E. for the C.S. gives
wc h1 + ws h2 = wc h3 + ws h4
or
ws (h2 – h4) = wc(h3 – h1)
Here the K.E. and P.E. terms are considered small, there is no external work done, and
energy exchange in the form of heat is confined only between the two fluids, i.e.
there is no external heat interaction or heat loss.
ws
2
2
Exhaust steam
Cooling
water in
wc
1
3
1
3
Cooling
water out
wc
4
4
Condensate
Fig. 5.7 Steam Condenser
Figure 5.8 shows a steam desuperheater where the temperature of the superheated steam is reduced by spraying water. If w1, w2, and w3 are the mass flow rates
of the injected water, of the steam entering, and of the steam leaving, respectively,
and h1, h2, and h3 are the corresponding enthalpies, and if K.E. and P.E. terms are
neglected as before, the S.F.E.E. becomes
w1h1 + w2h2 = w3h3
First Law Applied to Flow Processes
95
and the mass balance gives
w1 + w2 = w3
High temperature steam
w2
w1
Water
C.S.
3
w3
3
Low
temperature
steam
Steam Desuperheater
Fig. 5.8
5.5 COMPARISON OF S.F.E.E. WITH EULER AND BERNOULLI
EQUATIONS
The steady flow energy equation (Eq. 5.8) can be written as
_
dQ
dWx
V 2 - V12
= (h2 – h1) + 2
+ (Z2 – Z1 ) g +
dm
dm
2
In the differential form the S.F.E.E. becomes
_
d Q = dh + V d V + gdZ + dWx
(5.17)
_
_
_
where d Q and d Wx refer to unit mass of the substance. Since h = u + pv and d Q = dv
+ pdv (for a quasi-static path involving only pdv-work), Eq. (5.17) can be written as
_
du + pdv = du + pdv + vdp + VdV + gdZ + d Wx
For an inviscid frictionless fluid flowing through a pipe
vdp + VdV + gdZ = 0
(5.18)
This is the Euler equation. If we integrate between two Sections 1 and 2 of the
pipe
z z
2
2
vdp +
1
VdV +
1
z
2
gdZ = 0
1
For an incompressible fluid, v = constant
\
v(p2 – p1) +
V2
V22
– 1 + g (Z2 – Z1) = 0
2
2
(5.19)
Since the specific volume v is the reciprocal of the density r, we have
p1 V12
p
V2
+ Z1g = 2 + 2 + Z2g
+
r
r
2
2
(5.20)
96
Engineering Thermodynamics
p V2
+
+ Zg = constant
r
2
or
(5.21)
This is known as the Bernoulli equation, which is valid for an inviscid incompressible fluid. It can also be expressed in the following form
FG
H
D pv +
IJ
K
V2
+ gZ = 0
2
(5.22)
where v is constant and D (...) means ‘increase in ...
The S.F.E.E. as given by Eq. (5.8) or Eq. (5.17) can be written with (u + pv)
substituted for h, as follows:
FG
H
Q – Wx = D u + pv +
V2
+ gZ
2
IJ
K
(5.23)
A comparison of Eqs (5.22) and (5.23) shows that they have several terms in
common. However, while the Bernoulli equation is restricted to frictionless incompressible fluids, the S.F.E.E. is not, and is valid for viscous compressible fluids as
well. The Bernoulli equation is, therefore, a special limiting case of the more general
steady flow energy equation.
5.6
VARIABLE FLOW PROCESSES
Many flow processes, such as filling up and evacuating gas cylinders, are not
steady. Such processes can be analyzed by the control volume technique. Consider
a device through which a fluid is flowing under non-steady state conditions
(Fig. 5.9). The rate at which the mass of fluid within the control volume is accumulated is equal to the net rate of mass flow across the control surface, as given below
C.S.
dQ
dt
C.V.
w2
w1
dwx
dt
Fig. 5.9
Variable Flow Process
dmV
dm1 dm2
= w1 – w2 =
dt
dt
dt
where mV is the mass of fluid within the control volume at any instant.
Over any finite period of time
DmV = Dm1 – Dm2
(5.24)
(5.25)
97
First Law Applied to Flow Processes
The rate of accumulation of energy within the control volume is equal to the net
rate of energy flow across the control surface. If Ev is the energy of fluid within the
control volume at any instant
Rate of energy increase = Rate of energy inflow – Rate of energy outflow
_
dEV
dQ
V12
+ Z1g +
= w1 h1 +
dt
dt
2
FG
H
IJ
K
_
F
I
dW
V
– w Gh +
H 2 + Z gJK - dt
F mV + mgZIJ
E = GU +
H 2
K
2
2
2
x
2
(5.26)
2
2
Now
V
V
where m is the mass of fluid in the control volume at any instant.
FG
H
IJ
K
dEV
d
mV 2
=
+ mgZ
U+
dt
dt
2
V
\
_
dQ
dm1
V12
+ Z1g
+
dt
dt
2
_
d Wx
dm2
V2
– h2 + 2 + Z2 g
dt
dt
2
FG
H
IJ
K
= h1 +
FG
H
IJ
K
(5.27)
Figure 5.10 shows all these energy flux quantities. For any finite time interval, Eq.
(5.27) becomes
dwx
dt
dQ
dt
d
mV
U+
dt
2
–2
w1 h1 + V 1
2
+ Z1g
–2
+ mgZ
V
–2
w2 h2 + V 2
2
+ Z2g
C.S.
Fig. 5.10
Energy Fluxes in an Unsteady System
FG h + V + Z gIJ dm
H 2 K
F V + Z gIJ dm
– Gh +
H 2 K
DEV = Q – Wx +
z
2
1
z
1
1
2
2
2
2
2
1
(5.28)
98
Engineering Thermodynamics
Equation (5.26) is the general energy equation. For steady flow,
d EV
= 0,
dt
and the equation reduces to Eq. (5.7). For a closed system w1 = 0, w2 = 0, then from
Eq. (5.26),
_
_
d EV
dQ
d Wx
=
–
dt
dt
dt
_
_
_
_
or
dEV = d Q – d Wx or, d Q = dE + d Wx
as obtained earlier.
5.7
EXAMPLE OF A VARIABLE FLOW PROBLEM
Variable flow processes may be analyzed either by the system technique or the
control volume technique, as illustrated as follows.
Consider a process in which a gas bottle is filled from a pipeline (Fig. 5.11). In the
beginning the bottle contains gas of mass m1 at state p1, t1, v1, h1 and u1. The valve
is opened and gas flows into the bottle till the mass of gas in the bottle is m2 at state
p2, t2, v2, h2 and u2. The supply to the pipeline is very large so that the state of gas
in the pipeline is constant at pp, tp, vp, hp, up, and Vp.
Envelope of System boundary
Pipe line
System boundary
(Envelope)
C.S.
Tube
Bottle
W
Valve
Q
Fig. 5.11 Bottle-filling Process
System Technique Assume an envelope (which is extensible) of gas in the
pipeline and the tube which would eventually enter the bottle, as shown in Fig. 5.11.
Energy of the gas before filling
E1 = m1u1 + (m2 – m1)
FV +u I
GH 2 JK
2
p
p
where (m2 – m1) is the mass of gas in the pipeline and tube which would enter the
bottle.
Energy of the gas after filling
E2 = m2u2
LM
MN
b
DE = E2 – E1 = m2u2 – m1u1 + m2 - m1
O
gFGH V2 + u IJK PP
Q
2
p
p
(5.29)
99
First Law Applied to Flow Processes
The P.E. terms are neglected. The gas in the bottle is not in motion, and so the K.E.
terms have been omitted.
Now, there is a change in the volume of gas because of the collapse of the
envelope to zero volume. Then the work done
W = pp(V2 – V 1) = pp[0 – (m2 – m1)vp]
= – (m2 – m1)ppvp
\ Using the first law for the process
Q = DE + W
F V + u I – (m – m )p v
GH 2 JK
FV + h I
= m u – m u – (m – m ) G
H 2 JK
2
p
= m2u2 – m1u1 – (m2 – m1)
2
p
1
2
p
2 2
1 1
2
1
p
p p
(5.30)
which gives the energy balance for the process.
Control Volume Technique Assume a control volume bounded by a control
surface, as shown in Fig. 5.11. Applying the energy Eq. (5.27) to this case, the
following energy balance may be written on a time rate basis
_
Vp2 dm
d EV
dQ
=
+ hp +
dt
dt
2 dt
F
GH
I
JK
Since hp and Vp are constant, the equation is integrated to give for the total
process
F
GH
DEV = Q + h p +
I (m – m )
J
2 K
Vp2
2
1
Now
DEV = U2 – U1 = m2u2 – m1u1
\
F V I (m – m )
Q = m u – m u – Gh +
H 2 JK
2
p
2 2
1 1
p
2
1
This equation is the same as Eq. (5.30).
5.8 DISCHARGING AND CHARGING A TANK
Let us consider a tank discharging a fluid into a supply line (Fig. 5.12). Since
_
d Wx = 0 and dmin = 0, applying first law to the control volume,
FG
H
IJ
K
_
V2
dUV = d Q + h +
+ gz
dmout
2
out
_
Assuming K.E. and P.E. of the fluid to be small and d Q = 0
d(mu) = hdm
mdu + udm = udm + pv dm
(5.31)
100
Engineering Thermodynamics
Supply Line
Valve
C.S.
C.V.
Fig. 5.12 Charging and Discharging a Tank
du
dm
=
pv
m
Again
(5.32)
V = vm = const.
vdm + mdv = 0
dm
dv
=–
m
v
or
(5.33)
From euations (5.32) and 5.33),
du
dv
=–
v
pv
d(u + pv) = 0
_
dQ =0
or
which shows that the process is adiabatic and quasi-static.
For charging the tank
z
(hdm)in = DUV = m2u2 – m1u1
(5.34)
mphp = m2u2 – m1u1
where the subscript p refers to the constant state of the fluid in the pipeline. If the
tank is initially empty, m1 = 0.
mphp = m2u2
Since
mp = m2
hp = u2
101
First Law Applied to Flow Processes
If the fluid is an ideal gas, the temperature of the gas in the tank after it is charged is
given by
cpTp = cvT2
or
T2 = g Tp
(5.35)
Solved Examples
Example 5.1 Air flows steadily at the rate of 0.5 kg/s through an air compressor,
entering at 7 m/s velocity, 100 kPa pressure, and 0.95 m3 /kg volume, and leaving
at 5 m/s, 700 kPa, and 0.19 m3/kg. The internal energy of the air leaving is
90 kJ/kg greater than that of the air entering. Cooling water in the compressor
jackets absorbs heat from the air at the rate of 58 kW. (a) Compute the rate of shaft
work input to the air in kW. (b) Find the ratio of the inlet pipe diameter to outlet
pipe diameter.
Solution
Figure 5.13 shows the details of the problem.
V2 = 5 m/s
p2 = 700 kPa
v2 = 0.19 m3/kg
Wx
Air Compressor
V1 = 7 m/s
p1 = 100 kPa
v1 = 0.95 m3/kg
Q = - 58 kW
u2 = (u1 + 90 ) kJ / kg
C.S.
Fig. 5.13
(a) Writing the steady flow energy equation, we have
_
_
F
I
V
F
I
dQ
dW
V
+
+
+
u
p
Z
g
v
w Gu + p v +
+ Z gJ +
JK + dt
2
2
H
K dt = w GH
_
_
F
I
dW
dQ
V -V
= – w G bu - u g + b p v - p v g +
+ b Z - Z gg J
dt
2
H
K dt
_
dW
kg L kJ
kJ
= – 0.5
90
+ a7 ¥ 0.19 - 1 ¥ 0.95f 100
M
dt
s N kg
kg
d5 - 7 i ¥ 10 kJ + 0OP – 58 kW
+
2
kg
PQ
2
2
2
1
1
\
\
1 1
1
x
2
1
2 2
2
2 2
2
2
2
1
1 1
x
2
2
-3
= – 0.5 [90 + 38 – 0.012] kJ/s – 58 kW
= – 122 kW
Rate of work input is 122 kW.
2
2
1
x
102
Engineering Thermodynamics
(b) From mass balance, we have
w=
A1V1 A2 V2
=
v1
v2
\
0.95 5
A1
v V
= 1◊ 2 =
¥ = 3.57
A2
v 2 V1 0.19 7
\
d1
= 3.57 = 1.89
d2
Example 5.2 In a steady flow apparatus, 135 kJ of work is done by each kg of
fluid. The specific volume of the fluid, pressure, and velocity at the inlet are
0.37 m3/kg, 600 kPa, and 16 m/s. The inlet is 32 m above the floor, and the discharge pipe is at floor level. The discharge conditions are 0.62 m3/kg, 100 kPa,
and 270 m/s. The total heat loss between the inlet and discharge is 9 kJ/kg of fluid.
In flowing through this apparatus, does the specific internal energy increase or
decrease, and by how much?
Solution Writing the steady flow energy equation for the control volume, as
shown in Fig. 5.14.
C.S.
W = 135 kJ
v1 = 0.37 m3 / kg
p1 = 600 kPa
v2 = 0.62 m3 / kg
p2 = 100 kPa
C.V.
V1 = 16 m /s
V2 = 270 m/s
Z2 = 0
Z1 = 32 m
Q = - 9.0 kJ
Fig. 5.14
\
_
_
d Wx
dQ
V12
V22
u1 + p1v1 =
+ Z1g +
= u 2 + p2 v 2 +
+ Z 2g +
dm
dm
2
2
_
_
d Wx
dQ
V22 - V12
u1 – u2 = (p2v2 – p1v1) +
+ (Z2 – Z1)g +
–
dm
2
dm
d270 - 16 i ¥ 10
2
= (1 ¥ 0.62 – 6 ¥ 0.37) ¥ 102 +
–3
+ (– 32 ¥ 9.81 ¥ 10 ) + 135 – (– 9.0)
= – 160 + 36.45 – 0.314 + 135 + 9
= 20.136 kJ/kg
Specific internal energy decreases by 20.136 kJ.
2
2
-3
103
First Law Applied to Flow Processes
Example 5.3 In a steam power station, steam flows steadily through a 0.2 m
diameter pipeline from the boiler to the turbine. At the boiler end, the steam
conditions are found to be: p = 4 MPa, t = 400°C, h = 3213.6 kJ/kg, and
v = 0.073 m3/kg. At the turbine end, the conditions are found to be: p = 3.5 MPa,
t = 392°C, h = 3202.6 kJ/kg, and v = 0.084 m3/kg. There is a heat loss of 8.5 kJ/kg
from the pipeline. Calculate the steam flow rate.
Solution Writing the steady flow energy equation for the control volume as
shown in Fig. 5.15.
_
_
d Wx
dQ
V12
V22
h1 +
+ Z 1g +
= h2 +
+ Z 2g +
dm
dm
2
2
Here, there is no change in datum, so change in potential energy will be zero.
C.S.
Turbine
Boiler
C.V.
dQ
dm
= - 8.5 kJ / kg
Fig. 5.15
A1V1 A2 V2
=
v1
v2
Now
\
and
V2 =
A1V1 v2 v2
0.084
◊
=
◊ V1 =
V1 = 1.15 V1
0.073
v1 A2 v1
_
d Wx
=0
dm
h1 +
\
_
V12 d Q
V2
+
= h2 + 2
dm
2
2
dV - V i ¥ 10 = h – h + d_Q
2
2
2
1
2
-3
1
2
dm
= 3213.6 – 3202.6 + (– 8.5) = 2.5 kJ/kg
V21 (1.152 – 12) = 5 ¥ 103
V21 = 15,650 m2/s2
\
\
V1 = 125.1 m/s
p
2
¥ 0.2 m 2 ¥ 125.1 m /s
A1V1
4
Mass flow rate w =
=
= 53.8 kg/s
0.073 m 3 / kg
v1
a f
104
Engineering Thermodynamics
Example 5.4 A certain water
heater operates under steady flow
conditions receiving 4.2 kg/s of water at 75°C temperature, enthalpy
313.93 kJ/kg. The water is heated
by mixing with steam which is supplied to the heater at temperature
100.2°C and enthalpy 2676 kJ/kg.
The mixture leaves the heater as liquid water at temperature 100°C and
enthalpy 419 kJ/kg. How much
steam must be supplied to the heater
per hour?
w1
Water
C.S.
w2
Steam
Water heater
Mixture
Fig. 5.16
Solution By mass balance across
the control surface (Fig. 5.16)
w1 + w2 = w3
By energy balance
FG V + Z gIJ + d_ Q + w FG h + V + Z gIJ
H 2 K dt H 2 K
F V + Z gI + d_ W
= w Gh +
H 2 JK dt
w1 h1 +
2
1
2
2
2
3
3
3
2
1
2
2
x
3
By the nature of the process, there is no shaft work. Potential and kinetic energy
terms are assumed to balance zero. The heater is assumed to be insulated. So the
steady flow energy equation reduces to
w1h1 + w2h2 = w3h3
4.2 ¥ 313.93 + w2 ¥ 2676 = (4.2 + w2) 419
\
w2 = 0.196 kg/s = 705 kg/h
Example 5.5 Air at a temperature of 15°C passes through a heat exchanger at
a velocity of 30 m/s where its temperature is raised to 800°C. It then enters a
turbine with the same velocity of 30 m/s and expands until the temperature falls to
650°C. On leaving the turbine, the air is taken at a velocity of 60 m/s to a nozzle
where it expands until the temperature has fallen to 500°C. If the air flow rate is
2 kg/s, calculate (a) the rate of heat transfer to the air in the heat exchanger,
(b) the power output from the turbine assuming no heat loss, and (c) the velocity
at exit from the nozzle, assuming no heat loss. Take the enthalpy of air as h = cpt,
where cp is the specific heat equal to 1.005 kJ/kg K and t is the temperature.
Solution As shown in Fig. 5.17, writing the S.F.E.E. for the heat exchanger and
eliminating the terms not relevant,
105
First Law Applied to Flow Processes
Heat exchanger
WT
Q
Turbine
t1 = 15∞C, t2 = 800∞C
V1 = 30 m/s, V2 = 30 m/s
t3 = 650∞C, V3 = 60 m/s
t4 = 500∞C, V4 = ?
Nozzle
Fig. 5.17
FG
H
w h1 +
IJ + Q
K
V12
+ Z1g
2
\
1– 2 = w
FG h + V + Z gIJ + W
H 2 K
2
2
2
2
1– 2
wh1 + Q1 –2 = wh2
\
Q1–2 = w(h2 – h1) = w cp(t2 – t1 )
= 2 ¥ 1.005 (800 – 15)
= 2.01 ¥ 785
= 1580 kJ/s
Energy equation for the turbine gives
w
FG V + h IJ = wh + w V + W
2
H2 K
2
2
2
3
2
3
T
V22 - V32
+ (h2 – h3) = WT/w
2
-3
d30 – 60 i ¥ 10 + 1.005 (800 – 650) = W /w
2
2
T
2
\
WT
= 1.35 + 150.75
w
= 149.4 kJ/kg
\
WT = 149.4 ¥ 2 kJ/s
= 298.8 kW
Writing the energy equation for the nozzle
V33
V2
+ h3 = 4 + h4
2
2
V42 - V32
= cp (t3 – t4)
2
106
Engineering Thermodynamics
V24 – V23 = 1.005 (650 – 500) ¥ 2 ¥ 103
= 301.50 ¥ 103 m2/s2
V24 = 30.15 ¥ 104 + 0.36 ¥ 104
= 30.51 ¥ 104 m2/s2
\ Velocity at exit from the nozzle
V4 = 554 m/s
Example 5.6 In a gas turbine the gas enters at the rate of 5 kg/s with a velocity
of 50 m/s and enthalpy of 900 kJ/kg and leaves the turbine with a velocity of
150 m/s and enthalpy of 400 kJ/kg. The loss of heat from the gases to the
surroundings is 25 kJ/kg. Assume for gas R = 0.285 kJ/kg K and cp = 1.004 kJ/kgK
and the inlet conditions to be at 100 kPa and 27°C. Determine the power output of
the turbine and the diameter of the inlet pipe.
Solution
Steady flow energy equation for the C.V. gives
C.V.
1
Q
w
W
Turbine
2
w
Fig. 5.18
_
_
F
I
F
I
dQ
dW
V
V
w Gh +
H 2 + gZ JK + dt = w GH h + 2 + gZ JK + dt
_
_
L
dW
V -V O dQ
= w Mbh - h g +
+
P dt
dt
2 Q
N
LF
IO
50 - 150
= 5 MG 900 - 400 +
¥ 10 J P – 25 ¥ 5
2
K PQ
MNH
2
1
2
2
1
\
1
2
1
1
2
2
2
2
2
2
= 2325 kW
Using ideal gas equation of state at pipe inlet,
∑
p1 V1 = w1 R T1
∑
V = volume flow rate at inlet =
5 ¥ 0.285 ¥ 300 ¥ 10 3
100 ¥ 10 3
= 4.275 m3/s
-3
2
107
First Law Applied to Flow Processes
4.275
50
Inlet area,
A1 =
\
p 2
D1
4
D1 = 0.33 m or 33 cm
= 0.086 m2 =
Example 5.7 The air speed of a turbojet engine in flight is 270 m/s. Ambient air
temperature is – 15°C. Gas temperature of outlet of nozzle is 600°C. Corresponding enthalpy values for air and gas are respectively 260 and 912 kJ/kg. Fuel-air
ratio is 0.0190. Chemical energy of the fuel is 44.5 MJ/kg. Owing to incomplete
combustion 5% of the chemical energy is not released in the reaction. Heat loss
from the engine is 21 kJ/kg of air. Calculate the velocity of the exhaust jet.
Solution Energy equation for the turbojet engine (Fig. 5.19) gives
F
GH
I + w E + Q = w Fh + V + E I
JK
GH 2 JK
FG 260 + 270 ¥ 10 IJ + 0.0190 ¥ 44500 – 21
2
H
K
F V ¥ 10 + 0.05 0.019 ¥ 44500I
= 1.0190 G 912 +
JK
2
1.019
H
F V ¥ 10 + 42I
260 + 36.5 + 845 – 21 = 1.019 G 912 +
JK
2
H
wa ha +
Va2
2
2
a
f
f
g
g
g
-3
2
2
g
-3
2
g
Vg2
\
2
-3
= 156 ¥ 103 m2/s2
Vg =
312
. ¥ 100 m/s
Velocity of exhaust gas, Vg = 560 m/s
Fuel
Wf
Ef
Q
C.S.
3
Air
Wa
3
Wg
Exhaust
Fig. 5.19
Example 5.8 In a reciprocating engine, the mass of gas occupying the
clearance volume is mc kg at state p1, u1, v1 and h1. By opening the inlet valve, mf
108
Engineering Thermodynamics
kg of gas is taken into the cylinder, and at the conclusion of the intake process the
state of the gas is given by p2, u2, v2, h2. The state of the gas in the supply pipe is
constant and is given by pp, up, vp, hp, V p. How much heat is transferred between the
gas and the cylinder walls during the intake process?
Solution Let us consider the control volume as shown in Fig. 5.20. Writing the
energy balance on a time rate basis
_
_
Vp2 dm
d EV
dQ
dW
=
–
+ hp +
dt
dt
dt
2 dt
F
GH
Valve
I
JK
C.S.
Gas inlet
Q
Fig. 5.20
With hp and Vp being constant, the above equation can be integrated to give for
the total process
F
GH
DEV = Q – W + hp +
Im
J
2 K
Vp2
f
Now
DEv = U2 – U1 = (me + mf) u2 – meu1
\
Q = (mc + mf)u2 – mcu1 – mf hp +
Example 5.9
F
GH
I +W
J
2 K
Vp2
The internal energy of air is given by
u = u0 + 0.718 t
where u is in kJ/kg, u0 is any arbitrary value of u at 0°C, kJ/kg, and t is the
temperature in °C. Also for air, pv = 0.287 (t + 273), where p is in kPa and v is in
m3/kg.
A mass of air is stirred by a paddle wheel in an insulated constant volume tank.
The velocities due to stirring make a negligible contribution to the internal energy of the air. Air flows out through a small valve in the tank at a rate controlled
to keep the temperature in the tank constant. At a certain instant the conditions
are as follows: tank volume 0.12 m3, pressure 1 MPa, temperature 150°C, and
power to paddle wheel 0.1 kW. Find the rate of flow of air out of the tank at this
instant.
First Law Applied to Flow Processes
Solution
109
Writing the energy balance for the control volume as shown in Fig. 5.21
_
dW
d EV
dm
=
- hp
dt
dt
dt
d i
Since there is no change in internal energy of air in the tank,
_
dW
dm
=
hp ◊
dt
dt
W = 0.1 kW
where
h p = u + pv.
Let
u = 0 at t = 0 K = – 273°C
u = u0 + 0.718 t
0 = u0 + 0.718 (– 273)
Tank
u 0 = 0.718 ¥ 273 kJ/kg
Valve
At t°C
C.S.
u = 0.718 ¥ 273 + 0.718 t
= 0.718 (t + 273) kJ/kg
Fig. 5.21
h p = 0.718 (t + 273) + 0.287 (t + 273)
or
h p = 1.005 (t + 273)
At 150°C
hp = 1.005 ¥ 423
\
= 425 kJ/kg
_
dm
1 dW
=
dt
h p dt
=
0.1 kJ/s
= 0.236 ¥ 10–3 kg/s
425 kJ / kg
= 0.845 kg/h
This is the rate at which air flows out of the tank.
Example 5.10 A well-insulated vessel of volume V contains a gas at pressure p0
and temperature t0. The gas from a main at a uniform temperature t1 is pumped into
∑
the vessel and the inflow rate decreases exponentially with time according to m =
∑
m 0 e–at, where a is a constant. Determine the pressure and temperature of the gas
in the vessel as a function of time. Neglect the K.E. of the gas entering the vessel
and assume that the gas follows the relation
pv = RT, where T = t + 273
and its specific heats are constant.
(i) If the vessel was initially evacuated, show that the temperature inside the
vessel is independent of time.
110
Engineering Thermodynamics
(ii) Determine the charging time when the pressure inside the vessel reaches that
of the main.
∑
Since the vessel is well-insulated, Q = 0 and there is no external work
Solution
∑
transfer, W = 0. Therefore,
∑
dEV
dm
= h1
m = h1 m 0 e - at
dt
dt
where h1 is the enthalpy of the gas in the main
On integration,
∑
h1 m 0
(1 – e–at )
a
where E0 is the initial energy of the vessel at the beginning of the charging
process, i.e. E = E0 at t = 0. Neglecting K.E. and P.E. changes, by energy balance
E = E0 +
∑
m0
Mu = M0u0 +
(1 – e–at) (u1 + p1v1 )
a
(1)
Again,
∑
dm
= m 0 e - at
dt
On integration,
m 0 1 - e - at
∑
M = M0 +
d
i
a
where M0 is the initial mass of the gas. Eliminating M from Eqs. (1) and (2),
R| m
U
S| M + a d1 - e i|V| u – M u
T
W
∑
- at
0
0 0
0
∑
m0
=
(1 – e–at) (u1 + RT1)
a
∑
M0cv(T – T0) =
m0
(1 – a–at {c v (T1 – T) + RT1}
a
∑
\
(
T=
)
m0
1 - e- at c pT1
a
Ï
¸
m0
Ô
- at Ô
M
e
+
1
Ì 0
˝ cv
a
ÔÓ
Ô˛
M 0 cvT0 +
∑
(
)
Ï
¸
MRT
R Ô
m0
Ô
=
1 - e- at c pT1 ˝
Ì M 0 cvT0 +
V
Vcv Ô
a
Ô˛
Ó
∑
p=
(
)
(2)
First Law Applied to Flow Processes
111
∑
m0 R
(1 – e–at) g T1
aV
The above two equations show the temperature and pressure of the gas in the
vessel as functions of time.
(i) If M0 = 0, T = g T1, i.e. the temperature inside the vessel becomes independent
of time and is equal to g T1 throughout the charging process.
(ii) The charging process will stop when pressure inside the vessel reaches that
of the main. The charging time can be found by setting p = p1 in the pressure
relation
= p0 +
∑
p1 – p0 =
∑
m 0 Rg T1 m 0 R - at
e g T1
aV
aV
By rearrangement,
∑
eat =
\
t=
F
H
a f
I b
g
K
m 0 Rg T1 / aV
m0 Rg T1
- p1 - p0
aV
LM b
N
1
aV
ln 1 – p1 – p0
a
m0 Rg T1
g
OP
Q
Summary
The first law may be applied to flow processes. A control surface is the
boundary of a control volume which is a fixed region in space upon which
attention is concentrated in the analysis of a problem. A steady flow process
is a process in which all conditions within the control volume remain constant
with time. For a simple steady flow process involving one mass stream entering
and one mass stream leaving the control volume, the following equations
hold:
Mass balance or continuity equation
w=
AV
AV
1 1
= 2 2
v1
v2
Energy balance or steady flow energy equation (SFEE) written on a unit mass
basis
_
_
dQ
dQ
V2
V2
u1 + p1v1 + 1 + Z1g +
= u2 + p2v 2 + 2 + gZ2 +
dm
dm
2
2
or on time basis
_
_
Ê
ˆ dQ
Ê
ˆ d wx
V2
V2
w1 Á h1 + 1 + gZ1˜ +
= w2 Á h2 + 2 + gZ 2 ˜ +
dt
dt
2
2
Ë
¯
Ë
¯
112
Engineering Thermodynamics
For variable flow problems energy equations may be set up by means of a
system technique or control volume technique. In these cases account must
be taken not only of mass and energy quantities crossing the control surface
but also of accumulations within the control volume.
Review Questions
5.1 Explain the system approach and the control volume approach in the analysis
of a flow process.
5.2 What is a steady flow process?
5.3 Write the steady flow energy equation for a single stream entering and a
single stream leaving a control volume and explain the various terms in it.
5.4 Give the differential form of the S.F.E.E.
5.5 Under what conditions does the S.F.E.E. reduce to Euler’s equation?
5.6 How does Bernoulli’s equation compare with S.F.E.E.?
5.7 What will be the velocity of a fluid leaving a nozzle, if the velocity of approach
is very small?
5.8 Show that the enthalpy of a fluid before throttling is equal to that after throttling.
5.9 Write the general energy equation for a variable flow process.
5.10 What is the system technique in a bottle-filling process?
5.11 Explain the control volume technique in a variable flow process.
Problems
5.1 A blower handles 1 kg/s of air at 20°C and consumes a power of 15 kW.
The inlet and outlet velocities of air are 100 m/s and 150 m/s respectively.
Find the exit air temperature, assuming adiabatic conditions. Take cp of air is
1.005 kJ/kg-K.
Ans. 28.38°C
5.2 A turbine operates under steady flow conditions, receiving steam at the following state: pressure 1.2 MPa, temperature 188°C, enthalpy 2785 kJ/kg, velocity
33.3 m/s and elevation 3 m. The steam leaves the turbine at the following state:
pressure 20 kPa, enthalpy 2512 kJ/kg, velocity 100 m/s, and elevation 0 m. Heat
is lost to the surroundings at the rate of 0.29 kJ/s. If the rate of steam flow
through the turbine is 0.42 kg/s, what is the power output of the turbine in
kW?
Ans. 112.51 kW
5.3 A nozzle is a device for increasing the velocity of a steadily flowing stream. At
the inlet to a certain nozzle, the enthalpy of the fluid passing is 3000 kJ/kg and
the velocity is 60 m/s. At the discharge end, the enthalpy is 2762 kJ/kg. The
nozzle is horizontal and there is negligible heat loss from it. (a) Find the velocity at exict from the nozzle. (b) If the inlet area is 0.1 m2 and the specific volume
at inlet is 0.187 m3/kg, find the mass flow rate. (c) If the specific volume at the
nozzle exit is 0.498 m3/kg, find the exit area of the nozzle.
Ans. (a) 692.5 m/s, (b) 32.08 kg/s (c) 0.023 m2
113
First Law Applied to Flow Processes
5.4 In an oil cooler, oil flows steadily through a bundle of metal tubes submerged
in a steady stream of cooling water. Under steady flow conditions, the oil
enters at 90°C and leaves at 30°C, while the water enters at 25°C and leaves at
70°C. The enthalpy of oil at t°C is given by
h = 1.68 t + 10.5 ¥ 10–4 t2 kJ/kg
What is the cooling water flow required for cooling 2.78 kg/s of oil?
Ans. 1.47 kg/s
5.5 A thermoelectric generator consists of a series of semiconductor elements
(Fig. 5.22), heated on one side and cooled on the other. Electric current flow is
produced as a result of energy transfer as heat. In a particular experiment the
current was measured to be 0.5 amp and the electrostatic potential at (1) was
0.8 volt above that at (2). Energy transfer as heat to the hot side of the generator was taking place at a rate of 5.5 watts. Determine the rate of energy transfer
as heat from the cold side and the energy conversion efficiency.
Q1
Q1
Hot
P
N
P
N
P
N
+
Cold
Q2
Q2
e1 - e2
Fig. 5.22
Ans. Q2 = 5.1 watts, h = 0.073
5.6 A turbo compressor delivers 2.33 m3/s at 0.276 MPa, 43°C which is heated
at this pressure to 430°C and finally expanded in a turbine which delivers
1860 kW. During the expansion, there is a heat transfer of 0.09 MJ/s to the
surroundings. Calculate the turbine exhaust temperature if changes in kinetic
and potential energy are negligible.
Ans. 157°C
5.7 A reciprocating air compressor takes in 2 m3/min at 0.11 MPa, 20°C which it
delivers at 1.5 MPa, 111°C to an aftercooler where the air is cooled at constant
pressure to 25°C. The power absorbed by the compressor is 4.15 kW. Determine the heat transfer in (a) the compressor, and (b) the cooler. State your
assumptions.
Ans. – 0.17 kJ/s, – 3.76 kJ/s
5.8 In a water cooling tower air enters at a height of 1 m above the ground level
and leaves at a height of 7 m. The inlet and outlet velocities are 20 m/s and
30 m/s respectively. Water enters at a height of 8 m and leaves at a height of
0.8 m. The velocity of water at entry and exit are 3 m/s and 1 m/s respectively.
Water temperatures are 80°C and 50°C at the entry and exit respectively.
114
Engineering Thermodynamics
Air temperatures are 30°C and 70°C at the entry and exit respectively. The
cooling tower is well insulated and a fan of 2.25 kW drives the air through the
cooler. Find the amount of air per second required for 1 kg/s of water flow. The
values of cp of air and water are 1.005 and 4.187 kJ/kg K respectively.
Ans. 3.16 kg/s
5.9 Air at 101.325 kPa, 20°C is taken into a gas turbine power plant at a velocity of
140 m/s through an opening of 0.15 m2 cross-sectional area. The air is compressed heated, expanded through a turbine, and exhausted at 0.18 MPa,
150°C through an opening of 0.10 m2 cross-sectional area. The power output
is 375 kW. Calculate the net amount of heat added to the air in kJ/kg. Assume
that air obeys the law pv = 0.287 (t + 273), where p is the pressure in kPa, v is
the specific volume in m3/kg, and t is the temperature in °C. Take cp =
1.005 kJ/kg K.
Ans. 150.23 kJ/kg
5.10 A gas flows steadily through a rotary compressor. The gas enters the
compressor at a temperature of 16°C, a pressure of 100 kPa, and an enthalpy of
391.2 kJ/kg. The gas leaves the compressor at a temperature of 245°C, a pressure of 0.6 MPa, and an enthalpy of 534.5 kJ/kg. There is no heat transfer to or
from the gas as it flows through the compressor. (a) Evaluate the external work
done per unit mass of gas assuming the gas velocities at entry and exit to be
negligible. (b) Evaluate the external work done per unit mass of gas when the
gas velocity at entry is 80 m/s and that at exit is 160 m/s.
Ans. 143.3 kJ/kg, 152.9 kJ/kg
5.11 The steam supply to an engine comprises two streams which mix before entering the engine. One stream is supplied at the rate of 0.01 kg/s with an enthalpy
of 2952 kJ/kg and a velocity of 20 m/s. The other stream is supplied at the rate
of 0.1 kg/s with an enthalpy of 2569 kJ/kg and a velocity of 120 m/s. At the
exit from the engine the fluid leaves as two streams, one of water at the rate of
0.001 kg/s with an enthalpy of 420 kJ/kg and the other of steam; the fluid
velocities at the exit are negligible. The engine develops a shaft power of
25 kW. The heat transfer is negligible. Evaluate the enthalpy of the second exit
stream.
Ans. 2402 kJ/kg
5.12 The stream of air and gasoline vapour, in the ratio of 14 : 1 by mass, enters a
gasoline engine at a temperature of 30°C and leaves as combustion products
at a temperature of 790°C. The engine has a specific fuel consumption of
0.3 kg/kWh. The net heat transfer rate from the fuel-air stream to the jacket
cooling water and to the surroundings is 35 kW. The shaft power delivered by
the engine is 26 kW. Compute the increase in the specific enthalpy of the fuelair stream, assuming the changes in kinetic energy and in elevation to be
negligible.
Ans. – 1877 kJ/kg mixture
5.13 An air turbine forms part of an aircraft refrigerating plant. Air at a pressure
of 295 kPa and a temperature of 58°C flows steadily into the turbine with a
velocity of 45 m/s. The air leaves the turbine at a pressure of 115 kPa, a
temperature of 2°C, and a velocity of 150 m/s. The shaft work delivered by the
turbine is 54 kJ/kg of air. Neglecting changes in elevation, determine the magnitude and sign of the heat transfer per unit mass of air flowing. For air, take cp
= 1.005 kJ/kg K and the enthalpy h = cp t.
Ans. + 7.96 kJ/kg
First Law Applied to Flow Processes
115
5.14 In a turbomachine handling an incompressible fluid with a density of
1000 kg/m3 the conditions of the fluid at the rotor entry and exit are as given
below
Inlet
Exit
Pressure
1.15 MPa
0.05 MPa
Velocity
30 m/s
15.5 m/s
Height above datum
10 m
2m
If the volume flow rate of the fluid is 40 m3/s, estimate the net energy transfer from the fluid as work.
Ans. 60.3 MW
5.15 A room for four persons has two fans, each consuming 0.18 kW power, and
three 100 W lamps. Ventilation air at the rate of 80 kg/h enters with an enthalpy
of 84 kJ/kg and leaves with an enthalpy of 59 kJ/kg. If each person puts out
heat at the rate of 630 kJ/h determine the rate at which heat is to be removed by
a room cooler, so that a steady state is maintained in the room.
Ans. 1.92 kW
5.16 Air flows steadily at the rate of 0.4 kg/s through an air compressor,
entering at 6 m/s with a pressure of 1 bar and a specific volume of 0.85 m3/kg,
and leaving at 4.5 m/s with a pressure of 6.9 bar and a specific volume of
0.16 m3/kg. The internal energy of the air leaving is 88 kJ/kg greater than that
of the air entering. Cooling water in a jacket surrounding the cylinder absorbs
heat from the air at the rate of 59 W. Calculate the power required to drive the
compressor and the inlet and outlet cross-sectional areas.
Ans. 45.4 kW, 0.057 m2, 0.0142 m2
5.17 Two streams of air, one at 1 bar, 27°C and velocity of 30 m/s and the other at
5 bar, 227°C and 50 m/s velocity, mix in equal proportion in a chamber from
which heat at the rate of 100 kJ/kg is removed. The mixture is then passed
through an adiabatic nozzle. Find the velocity of the stream issuing out
of the nozzle. The temperature of air leaving the nozzle is 27°C, and its cp =
1.005 kJ/kgK.
Ans. 51.96 m/s
5.18 Steam flowing in a pipeline is at a steady state represented by pp , tp , up , v p, hp
and Vp. A small amount of the total flow is led through a small tube to an
evacuated chamber which is allowed to fill slowly until the pressure is equal to
the pipeline pressure. If there is no heat transfer, derive an expression for the
final specific internal energy in the chamber, in terms of the properties in the
pipeline.
5.19 The internal energy of air is given, at ordinary temperatures, by
u = u0 + 0.718 t
where u is in kJ/kg, u0 is any arbitrary value of u at 0°C, kJ/kg, and t is temperature in °C.
Also for air, pv = 0.287 (t + 273)
where p is in kPa and v is in m3/kg.
(a) An evacuated bottle is fitted with a valve through which air from the
atmosphere, at 760 mm Hg and 25°C, is allowed to flow slowly to fill the
bottle. If no heat is transferred to or from the air in the bottle, what will its
temperature be when the pressure in the bottle reaches 760 mm Hg?
Ans. 144.2°C
116
Engineering Thermodynamics
(b) If the bottle initially contains 0.03 m3 of air at 400 mm Hg and 25°C, what
will the temperature be when the pressure in the bottle reaches 760 mm
Hg?
Ans. 71.6°C
5.20 A pressure cylinder of volume V contains air at pressure p0 and temperature
T0. It is to be filled from a compressed air line maintained at constant pressure
p1 and temperature T1. Show that the temperature of the air in the cylinder
after it has been charged to the pressure of the line is given by
T=
g T1
p
T
1+ 0 g 1 -1
p1 T0
FG
H
IJ
K
5.21 A small reciprocating vacuum pump having the rate of volume displacement
V d is used to evacuate a large vessel of volume V. The air in the vessel is
maintained at a constant temperature T by energy transfer as heat. If the initial
and final pressures are p1 and p2 respectively, find the time taken for the
pressure drop and the necessary energy transfer as heat during evacuation.
Assume that for air, pV = mRT, where m is the mass and R is a constant, and u
is a function of T only.
LMAns. t = V ln p ; Q = b p - p gV OP
V
p
N
Q
1
d
1
2
2
[Hint: dm = – p(V d ◊ dt)/(RT) = V dp/(RT)].
5.22 A tank containing 45 kg of water initially at 45°C has one inlet and one exit
with equal mass flow rates. Liquid water enters at 45°C and a mass flow rate of
270 kg/h. A cooling coil immersed in the water removes energy at the ate of
7.6 kW. The water is well mixed by a paddle wheel with a power input of
0.6 kW. The pressures at inlet and exit are equal. Ignoring changes in KE and
PE, find the variation of water temperature with time.
Ans. T = 318 – 22 [1 – exp (– 6t)]
5.23 A rigid tank of volume 0.5 m3 is initially evacuated. A tiny hole develops in the
wall, and air from the surroundings at 1 bar, 21°C leaks in. Eventually, the
pressure in the tank reaches 1 bar. The process occurs slowly enough that
heat transfer between the tank and the surroundings keeps the temperature of
the air inside the tank constant at 21°C. Determine the amount of heat transfer.
Ans. 50 kJ
5.24 A well-insulated rigid tank with a volume of 10 m3 is connected to a large
steam line through which steam flows at 15 bar and 280°C. The tank is initially
evacuated. Steam is allowed to flow into the tank until the pressure inside is
15 bar. Calculate the amount of mass that flows into the tank.
Ans. 47.4 kg
6
Second Law of
Thermodynamics
6.1 QUALITATIVE DIFFERENCE BETWEEN HEAT AND WORK
The first law of thermodynamics states that a certain energy balance will hold when
a system undergoes a change of state or a thermodynamic process. But it does not
give any information on whether that change of state or the process is at all feasible
or not. The first law cannot indicate whether a metallic bar of uniform temperature
can spontaneously become warmer at one end and cooler at the other. All that the
law can state is that if this process did occur, the energy gained by one end would be
exactly equal to that lost by the other. It is the second law of thermodynamics which
provides the criterion as to the probability of various processes.
Spontaneous processes in nature occur only in one direction. Heat always flows
from a body at a higher temperature to a body at a lower temperature, water always
flows downward, time always flows in the forward direction. The reverse of these
never happens spontaneously. The spontaneity of the process is due to a finite driving potential, sometimes called the ‘force’ or the ‘cause’, and what happens is called
the ‘flux’, the ‘current’ or the ‘effect’. The typical forces like temperature gradient,
concentration gradient and electric potential gradient, have their respective conjugate fluxes of heat transfer, mass transfer, and flow of electric current. These transfer processes can never spontaneously occur from a lower to a higher potential.
This directional law puts a limitation on energy transformation other than that imposed by the first law.
Joule’s experiments (Article 4.1) amply demonstrate that energy, when
supplied to a system in the form of work, can be completely converted into heat
(work transfer Æ internal energy increase Æ heat transfer). But the complete conversion of heat into work in a cycle is not possible. So heat and work are not completely interchangeable forms of energy.
When work is converted into heat, we always have
W= Q
but when heat is converted into work in a complete closed cycle process
Q > W
118
Engineering Thermodynamics
The arrow indicates the direction of energy transformation. This is illustrated in
Fig. 6.1. As shown in Fig. 6.1(a), a system is taken from state 1 to state 2 by work
transfer W1 – 2 , and then by heat transfer Q2 – 1 the system is brought back from state
2 to state 1 to complete a cycle. It is always found that W1 – 2 = Q2 – 1 . But if the
system is taken from state 1 to state 2 by heat transfer Q1 – 2, as shown in Fig. 6.1(b),
then the system cannot be brought back from state 2 to state 1 by work transfer W2
– 1 . Hence, heat cannot be converted completely and continuously into work in a
cycle. Some heat has to be rejected. In Fig. 6.1(b), W2 – 3 is the work done and
Q3 – 1 is the heat rejected to complete the cycle. This underlies the work of Sadi
Carnot, a French military engineer, who first studied this aspect of energy transformation (1824). Work is said to be a high grade energy and heat a low grade energy.
The complete conversion of low grade energy into high grade energy in a cycle is
impossible.
Q2 – 1
1
2
1
1
W1 – 2 = Q2 – 1
W1 – 2
(a)
Fig. 6.1
6.2
2
Q1 – 2
1
W2 – 1
Q1 – 2 > W2 – 1
(b)
Qualitative Distinction between Heat and Work
CYCLIC HEAT ENGINE
For engineering purposes, the second law is best expressed in terms of the
conditions which govern the production of work by a thermodynamic system
operating in a cycle.
A heat engine cycle is a thermodynamic cycle in which there is a net heat transfer to the system and a net work transfer from the system. The system which executes a heat engine cycle is called a heat engine.
A heat engine may be in the form of mass of gas confined in a cylinder and
piston machine (Fig. 6.2a) or a mass of water moving in a steady flow through a
steam power plant (Fig. 6.2b).
In the cyclic heat engine, as represented in Fig. 6.2(a), heat Q1 is transferred to
the system, work WE is done by the system, work Wc is done upon the system, and
then heat Q2 is rejected from the system. The system is brought back to the initial
state through all these four successive processes which constitute a heat engine
cycle. In Fig. 6.2(b) heat Q1 is transferred from the furnace to the water in the boiler
to form steam which then works on the turbine rotor to produce work WT , then the
steam is condensed to water in the condenser in which an amount of heat Q2 is
rejected from the system, and finally work Wp is done on the system (water) to
pump it to the boiler. The system repeats the cycle.
The net heat transfer in a cycle to either of the heat engines
Qnet = Q1 – Q2
(6.1)
119
Second Law of Thermodynamics
Q1
WE
System
WC
(a)
Q2
H2O
Vapour
WT
Turbine
Furnace
Boiler
Q1
Condenser
H2O
Q2
Sea, River or
Atmosphere
Pump
(b)
Fig. 6.2
Wp
Cycle Heat Engine
(a) Heat Engine Cycle Performed by a Closed System Undergoing Four
Successive Energy Interactions with the Surroundings
(b) Heat Engine Cycle Performed by a Steady Flow System Interacting
with the Surroundings as Shown
and the net work transfer in a cycle
Wnet = WT – WP
(or
(6.2)
Wnet = WE – WC)
By the first law of thermodynamics, we have
 Q=  W
cycle
cycle
\
Qnet = Wnet
or
Q1 – Q2 = WT – WP
Figure 6.3 represents a cyclic heat engine in the
form of a block diagram indicating the various
energy interactions during a cycle. Boiler (B),
turbine (T), condenser (C), and pump (P), all
four together constitute a heat engine. A heat
engine is here a certain quantity of water undergoing the energy interactions, as shown, in cyclic operations to produce net work from a certain heat input.
The function of a heat engine cycle is to produce work continuously at the expense of heat
input to the system. So the net work Wnet and
(6.3)
H2O(l)
Q1
H2O(g)
B
WP
P
T
WT
C
H2O
H2O
Q2
Fig. 6.3
Cyclic Heat Engine with
Energy Interactions
Represented in a Block
Diagram
120
Engineering Thermodynamics
heat input Q1 referred to the cycle are of primary interest. The efficiency of a heat
engine or a heat engine cycle is defined as follows:
h=
Net work output of the cycle Wnet
=
Q1
Total heat input to the cycle
(6.4)
From Eqs (6.1), (6.2), (6.3) and (6.4),
h=
Wnet
W - WP
Q - Q2
= T
= 1
Q1
Q1
Q1
h=1–
Q2
Q1
(6.5)
This is also known as the thermal efficiency of a heat engine cycle. A heat engine
is very often called upon to extract as much work (net) as possible from a certain
heat input, i.e. to maximize the cycle efficiency.
6.3
ENERGY RESERVOIRS
A thermal energy reservoir (TER) is defined as a large body of infinite heat
capacity, which is capable of absorbing or rejecting an unlimited quantity of heat
without suffering appreciable changes in its thermodynamic coordinates. The
changes that do take place in the large body as heat enters or leaves are so very slow
and so very minute that all processes within it are quasi-static.
The thermal energy reservoir TERH from which heat Q1 is transferred to the
system operating in a heat engine cycle is called the source. The thermal energy
reservoir TERL to which heat Q2 is
rejected from the system during a
TERH
cycle is the sink. A typical source is
(Source)
a constant temperature furnace
where fuel is continuously burnt, and
Q1
a typical sink is a river or sea or the
WT
atmosphere itself.
B
WP
A mechanical energy reservoir
P
T
MER
(MER) is a large body enclosed by
C
Wnet
an adiabatic impermeable wall caCHE
Q2
pable of storing work as potential energy (such as a raised weight or
wound spring) or kinetic energy
TERL
(such as a rotating flywheel). All
(Sink)
processes of interest within an MER
are essentially quasi-static. An MER Fig. 6.4 Cyclic Heat Engine (CHE) with
Source and Sink
receives and delivers mechanical energy quasi-statically.
Figure 6.4 shows a cyclic heat engine exchanging heat with a source and a sink
and delivering Wnet in a cycle to an MER.
121
Second Law of Thermodynamics
6.4 KELVIN-PLANCK STATEMENT OF SECOND LAW
The efficiency of a heat engine is given by
h=
Wnet
Q
=1– 2
Q1
Q1
Experience shows that Wnet < Q1, since heat Q1 transferred to a system cannot be
completely converted to work in a cycle (Article 6.1). Therefore, h is less than
unity. A heat engine can never be 100% efficient. Therefore, Q2 > 0, i.e. there has
always to be a heat rejection. To produce net work in a thermodynamic cycle, a heat
engine has thus to exchange heat with two reservoirs, the source and the sink.
The Kelvin-Planck statement of the second law states: It is impossible for a heat
engine to produce net work in a complete cycle if it exchanges heat only with bodies at a single fixed temperature.
If Q2 = 0 (i.e. Wnet = Q1, or h = 1.00), the heat engine will produce net work in a
complete cycle by exchanging heat with only one reservoir, thus violating the
Kelvin-Planck statement (Fig. 6.5). Such a heat engine is called a perpetual motion
machine of the second kind, abbreviated to PMM2. A PMM2 is impossible.
A heat engine has, therefore, to exchange heat with two thermal energy
reservoirs at two different temperatures to produce net work in a complete cycle
(Fig. 6.6). So long as there is a difference in temperature, motive power (i.e. work)
can be produced. If the bodies with which the heat engine exchanges heat are of
finite heat capacities, work will be produced by the heat engine till the temperatures
of the two bodies are equalized.
Source at t1
t1
Q1
Q1
H.E
H.E
Wnet = Q1
Wnet
Q2
Q2 = O
Sink at t2
Fig. 6.5
6.5
A PMM2
Fig. 6.6
Heat Engine Producing Net
Work in a Cycle by Exchanging
Heat at Two Different
Temperatures
CLAUSIUS’ STATEMENT OF THE SECOND LAW
Heat always flows from a body at a higher temperature to a body at a lower temperature. The reverse process never occurs spontaneously.
122
Engineering Thermodynamics
Clausius’ statement of the second law gives: It is impossible to construct a device which, operating in a cycle, will produce no effect other than the transfer of
heat from a cooler to a hotter body.
Heat cannot flow of itself from a body at a lower temperature to a body at a
higher temperature. Some work must be expended to achieve this.
6.6 REFRIGERATOR AND HEAT PUMP
A refrigerator is a device which, operating in a cycle, maintains a body at a temperature lower than the temperature of the surroundings. Let the body A (Fig. 6.7)
be maintained at t2, which is lower than the ambient temperature t1. Even though A
is insulated, there will always be heat leakage Q2 into the body from the surroundings by virtue of the temperature difference. In order to maintain body A at the
constant temperature t 2, heat has to be removed from the body at the same rate at
which heat is leaking into the body. This heat (Q2) is absorbed by a working fluid,
called the refrigerant, which evaporates in the evaporator E1 at a temperature lower
than t2 absorbing the latent heat of vaporization from the body A which is cooled or
refrigerated (Process 4–1). The vapour is first compressed in the compressor C1
driven by a motor which absorbs work WC (Process 1–2), and is then condensed in
the condenser C2 rejecting the latent heat of condensation Q1 at a temperature higher
than that of the atmosphere (at t1) for heat transfer to take place (Process 2–3). The
condensate then expands adiabatically through an expander (an engine or turbine)
producing work WE, when the temperature drops to a value lower than t2 such that
heat Q2 flows from the body A to make the refrigerant evaporate (Process 3–4).
Such a cyclic device of flow through E1–C1–C2–E2 is called a refrigerator. In a
Atmosphere
t1
3
C2
Q1
Atmosphere
at t1
2
Condenser
Q1
Refrigerant
Compressor
3
E2
1
WE
C1
WE
Expander
E2
WC
Q2
Body A
t2
Q2
Body A
at t2
Q2
(a)
Fig. 6.7
1
Q2
E1
4
C1
E1
4
Evaporator
2
C2
(b)
A Cyclic Refrigeration Plant
WC
Second Law of Thermodynamics
123
refrigerator cycle, attention is concentrated on the body A. Q2 and W are of primary
interest. Just like efficiency in a heat engine cycle, there is a performance parameter
in a refrigerator cycle, called the coefficient of performance, abbreviated to COP,
which is defined as
COP =
\
[COP]ref =
Desired effect
Q
= 2
Work input
W
Q2
Q1 - Q2
(6.6)
A heat pump is a device which, operating in a cycle, maintains a body, say B
(Fig. 6.8), at a temperature higher than
the temperature of the surroundings. By
virtue of the temperature difference,
there will be heat leakage Q1 from the
body to the surroundings. The body will
be maintained at the constant temperature t1, if heat is discharged into the
body at the same rate at which heat
leaks out of the body. The heat is extracted from the low temperature reservoir, which is nothing but the atmosphere, and discharged into the high
temperature body B, with the expenditure of work W in a cyclic device
Fig. 6.8 A Cyclic Heat Pump
called a heat pump. The working fluid
operates in a cycle flowing through the
evaporator E1, compressor C1, condenser C2 and expander E2, similar to a refrigerator, but the attention is here focussed on the high temperature body B. Here Q1
and W are of primary interest, and the COP is defined as
Q
COP = 1
W
Q1
\
[COP]H.P. =
(6.7)
Q1 - Q2
From Eqs. (6.6) and (6.7), it is found that
[COP]H.P. = [COP]ref + 1
(6.8)
The COP of a heat pump is greater than the COP of a refrigerator by unity. Equation
(6.8) expresses a very interesting feature of a heat pump. Since
Q1 = [COP]H.P. W
= [COPref + 1] W
Q1 is always greater than W.
(6.9)
124
Engineering Thermodynamics
For an electrical resistance heater, if W is the electrical energy consumption,
then the heat transferred to the space at steady state is W only, i.e. Q1 = W.
A 1 kW electric heater can give 1 kW of heat at steady state and nothing more. In
other words, 1 kW of work (high grade energy) dissipates to give 1 kW of heat (low
grade energy), which is thermodynamically inefficient.
However, if this electrical energy W is used to drive the compressor of a heat
pump, the heat supplied Q1 will always be more than W, or Q1 > W. Thus, a heat
pump provides a thermodynamic advantage over direct heating.
For heat to flow from a cooler to a hotter body, W cannot be zero, and hence, the
COP (both for refrigerator and heat pump) cannot be infinity. Therefore,
W > 0, and COP < •.
6.7
EQUIVALENCE OF KELVIN-PLANCK AND CLAUSIUS
STATEMENTS
At first sight, Kelvin-Planck’s and Clausius’ statements may appear to be unconnected, but it can easily be shown that they are virtually two parallel statements of
the second law and are equivalent in all respects.
The equivalence of the two statements will be proved if it can be shown that the
violation of one statement implies the violation of the second, and vice versa.
(a) Let us first consider a cyclic heat pump P which transfers heat from a low
temperature reservoir (t2) to a high temperature reservoir (t 1) with no other effect,
i.e. with no expenditure of work, violating Clausius statement (Fig. 6.9).
Hot Reservoir at t1
Q1
Q1
W=0
H.P.
H.E.
P
Wnet = Q1 – Q2
E
Q1
Q2
Cold Reservoir at t2
Fig. 6.9 Violation of the Clausius Statement
Let us assume a cyclic heat engine E operating between the same thermal
energy reservoirs, producing Wnet in one cycle. The rate of working of the heat
engine is such that it draws an amount of heat Q1 from the hot reservoir equal to that
discharged by the heat pump. Then the hot reservoir may be eliminated and the heat
Q1 discharged by the heat pump is fed to the heat engine. So we see that the heat
Second Law of Thermodynamics
125
pump P and the heat engine E acting together constitute a heat engine operating in
cycles and producing net work while exchanging heat only with one body at a single
fixed temperature. This violates the Kelvin-Planck statement.
(b) Let us now consider a perpetual motion machine of the second kind (E)
which produces net work in a cycle by exchanging heat with only one thermal energy reservoir (at t1) and thus violates the Kelvin-Planck statement (Fig. 6.10).
t1
Q1
PMM2
(E)
H.E.
Q1 + Q2
W = Q1
H.P.
P
Q2 = 0
Q2
t2
Fig. 6.10 Violation of the Kelvin-Planck Statement
Let us assume a cyclic heat pump (P) extracting heat Q2 from a low temperature
reservoir at t2 and discharging heat to the high temperature reservoir at t1 with the
expenditure of work W equal to what the PMM2 delivers in a complete cycle. So E
and P together constitute a heat pump working in cycles and producing the sole
effect of transferring heat from a lower to a higher temperature body, thus violating
the Clausius statement.
6.8 REVERSIBILITY AND IRREVERSIBILITY
y
The second law of thermodynamics enables us to divide all processes into two
classes:
(a) Reversible or ideal process.
(b) Irreversible or natural process.
A reversible process is one which is perA
formed in such a way that at the conclusion of
the process, both the system and the surroundings may be restored to their initial states, without producing any changes in the rest of the universe. Let the state of a system be represented by
B
A (Fig. 6.11), and let the system be taken to state
B by following the path A–B. If the system and
x
also the surroundings are restored to their initial
Fig.
6.11
A
Reversible
Process
states and no change in the universe is produced,
126
Engineering Thermodynamics
then the process A–B will be a reversible process. In the reverse process, the system
has to be taken from state B to A by following the same path B–A. A reversible
process should not leave any trace or relic behind to show that the process had ever
occurred.
A reversible process is carried out infinitely slowly with an infinitesimal gradient, so that every state passed through by the system is an equilibrium state. So a
reversible process coincides with a quasi-static process.
Any natural process carried out with a finite gradient is an irreversible process.
A reversible process, which consists of a succession of equilibrium states, is an
idealized hypothetical process, approached only as a limit. It is said to be an asymptote to reality. All spontaneous processes are irreversible.
Time has an important effect on reversibility. If the time allowed for a process to
occur is infinitely large, even though the gradient is finite, the process becomes
reversible. However, if this time is squeezed to a finite value, the finite gradient
makes the process irreversible.
6.9 CAUSES OF IRREVERSIBILITY
The irreversibility of a process may be due to either one or both of the following:
(a) Lack of equilibrium during the process.
(b) Involvement of dissipative effects.
6.9.1
Irreversibility due to Lack of Equilibrium
The lack of equilibrium (mechanical, thermal or chemical) between the system and
its surroundings, or between two systems, or two parts of the same system, causes a
spontaneous change which is irreversible. The following are specific examples in
this regard:
(a) Heat Transfer through a Finite Temperature Difference A heat transfer process approaches reversibility as the temperature difference between two bodies approaches zero. We define a reversible heat transfer process as one in which
heat is transferred through an infinitesimal temperature difference. So to transfer a
finite amount of heat through an infinitesimal temperature difference would require
an infinite amount of time, or infinite area. All actual heat transfer processes are
through a finite temperature difference and are, therefore, irreversible, and greater
the temperature difference, the greater is the irreversibility.
We can demonstrate by the second law that heat transfer through a finite temperature difference is irreversible. Let us assume that a source at tA and a sink at tB
(t A > tB) are available, and let QA – B be the amount of heat flowing from A to B
(Fig. 6.12). Let us assume an engine operating between A and B, taking heat Q1
from A and discharging heat Q2 to B. Let the heat transfer process be reversed, and
QB – A be the heat flowing from B to A, and let the rate of working of the engine be
such that
Q2 = QB – A
127
Second Law of Thermodynamics
Source A, tA
Source A, tA
Q1
Q1
Wnet
E
QA – B
QB – A
Q2
Heat Transfer
Through a Finite
Temperature Difference
Q2
Sink B, tB
Sink B, tB
Fig. 6.12
Wnet
E
Fig. 6.13
Heat Transfer Through
a Finite Temperature
Difference is Irreversible
(Fig. 6.13). Then the sink B may be eliminated. The net result is that E produces
network W in a cycle by exchanging heat only with A, thus violating the KelvinPlanck statement. So the heat transfer process QA – B is irreversible, and QB – A is not
possible.
(b) Lack of Pressure Equilibrium within the Interior of the System or between the System and the Surroundings When there exists a difference in
pressure between the system and the surroundings, or within the system itself, then
both the system and its surroundings or the system alone, will undergo a change of
state which will cease only when mechanical equilibrium is established. The reverse of this process is not possible spontaneously without producing any other
effect. That the reverse process will violate the second law becomes obvious from
the following illustration.
(c) Free Expansion Let us consider an insulated container (Fig. 6.14) which is
divided into two compartments A and B by a thin diaphragm. Compartment A contains a mass of gas, while compartment B is completely evacuated. If the diaphragm is punctured,
Gas
Vacuum
the gas in A will expand into B until the pressures in A and B become equal. This is known as
free or unrestrained expansion. We can demonA
B
strate by the second law, that the process of free
expansion is irreversible.
To prove this, let us assume that free expansion
Diaphragm
Insulation
is reversible, and that the gas in B returns into A
Fig. 6.14 Free Expansion
with an increase in pressure, and B becomes evacu-
128
Engineering Thermodynamics
ated as before (Fig. 6.15). There is no other
Q=W
effect. Let us install an engine (a machine,
not a cyclic heat engine) between A and B, and
B
A
permit the gas to expand through the engine
Heat
source, t
from A to B. The engine develops a work
output W at the expense of the internal energy of the gas. The internal energy of the
Engine
gas (system) in B can be restored to its initial
W
value by heat transfer Q (= W) from a source.
Now, by the use of the reversed free expan- Fig. 6.15 Second Law Demonstrates
sion, the system can be restored to the initial
that Free Expansion is
state of high pressure in A and vacuum in B.
Irreversible
The net result is a cycle, in which we observe
that net work output W is accomplished by exchanging heat with a single reservoir. This
violates the Kelvin-Planck statement. Hence, free expansion is irreversible.
The same argument will hold if the compartment B is not in vacuum but at a
pressure lower than that in compartment A (case b).
6.9.2
Irreversibility due to Dissipative Effects
The irreversibility of a process may be due to the dissipative effects in which work
is done without producing an equivalent increase in the kinetic or potential energy
of any system. The transformation of work into molecular internal energy either of
the system or of the reservoir takes place through the agency of such phenomena as
friction, viscosity, inelasticity, electrical resistance, and magnetic hysteresis. These
effects are known as dissipative effects, and work is said to be dissipated.
(a) Friction Friction is always present in moving devices. Friction may be reduced by suitable lubrication, but it can never be completely eliminated. If this
were possible, a movable device could be kept in continual motion without violating either of the two laws of thermodynamics. The continual motion of a movable
device in the complete absence of friction is known as perpetual motion of the third
kind.
That friction makes a process irreversFlywheel
Brake block
ible can be demonstrated by the second law.
Let us consider a system consisting of a flywheel and a brake block (Fig. 6.16). The
flywheel was rotating with a certain rpm,
F
and it was brought to rest by applying the
friction brake. The distance moved by the
brake block is very small, so work transfer
is very nearly equal to zero. If the braking
System boundary
process occurs very rapidly, there is little
heat transfer. Using suffix 2 after braking Fig. 6.16 Irreversibility due to
Dissipative Effect like
and suffix 1 before braking, and applying
Friction
the first law, we have
Q1 – 2 = E2 – E1 + W1 – 2
Second Law of Thermodynamics
129
0 = E2 – E1 + 0
\
E2 = E1
(6.10)
The energy of the system (isolated) remains constant. Since the energy may exist in
the forms of kinetic, potential, and molecular internal energy, we have
mV22
mV12
+ mZ 2 g = U1 +
+ mZ1g
2
2
Since the wheel is brought to rest, V2 = 0, and there is no change in P.E.
U2 +
mV12
(6.11)
2
Therefore, the molecular internal energy of the system (i.e., of the brake and the
wheel) increases by the absorption of the K.E. of the wheel. The reverse process,
i.e., the conversion of this increase in molecular internal energy into K.E. within the
system to cause the wheel to rotate is not possible. To prove it by the second law, let
us assume that it is possible, and imagine the following cycle with three processes:
U2 = U1 +
Process A Initially, the wheel and the brake are at high temperature as a result of
the absorption of the K.E. of the wheel, and the flywheel is at rest. Let the flywheel
now start rotating at a particular rpm at the expense of the internal energy of the
wheel and brake, the temperature of which will then decrease.
Process B Let the flywheel be brought to rest by using its K.E. in raising weights,
with no change in temperature.
Process C Now let heat be supplied from a source to the flywheel and the brake,
to restore the system to its initial state.
Therefore, the processes A, B, and C together constitute a cycle producing work
by exchanging heat with a single reservoir. This violates the Kelvin-Planck statement, and it will become a PMM2. So the braking process, i.e. the transformation
of K.E. into molecular internal energy, is irreversible.
(b) Paddle-Wheel Work Transfer
Work may be transferred into a system in
an insulated container by means of a paddle wheel (Fig. 6.17) which is also known
as stirring work. Here work transferred is dissipated adiabatically into an increase
in the molecular internal energy of the system. To prove the irreversibility of the
process, let us assume that the same amount of work is delivered by the system at
the expense of its molecular internal energy,
and the temperature of the system goes down
Insulation
(Fig. 6.18). The system is brought back to its
initial state by heat transfer from a source.
These two processes together constitute a
cycle in which there is work output and the
W
system exchanges heat with a single reservoir. It becomes a PMM2, and hence the disSystem
sipation of stirring work to internal energy is
Fig. 6.17 Adiabatic Work Transfer
irreversible.
130
Engineering Thermodynamics
Adiabatic
Diathermic
Q=W
W
Heat
source
Fig. 6.18 Irreversibility Due to Dissipation of Stirring Work into Internal Energy
(c) Transfer of Electricity through a Resistor
The flow of electric current
through a wire represents work transfer, because the current can drive a motor which
can raise a weight. Taking the wire or the
W
Resistor (system)
resistor as the system (Fig. 6.19) and writI
I
ing the first law
Q1 – 2 = U2 – U1 + W1 – 2
Here both W1 – 2 and Q1 – 2 are negative.
W1 – 2 = U2 – U1 + Q1 – 2
(6.12)
Q
Fig. 6.19
Irreversibility Due to
Dissipation of Electrical
Work into Internal Energy
A part of the work transfer is stored as an
increase in the internal energy of the wire
(to give an increase in its temperature), and
the remainder leaves the system as heat. At steady state, the internal energy and
hence the temperature of the resistor become constant with respect to time and
W1 – 2 = Q1 – 2
(6.13)
The reverse process, i.e. the conversion of heat Q1 – 2 into electrical work W1 – 2
of the same magnitude is not possible. Let us assume that this is possible. Then heat
Q1 – 2 will be absorbed and equal work W1 – 2 will be delivered. But this will become
a PMM2. So the dissipation of electrical work into internal energy or heat is irreversible.
6.10 CONDITIONS FOR REVERSIBILITY
A natural process is irreversible because the conditions for mechanical, thermal
and chemical equilibrium are not satisfied, and the dissipative effects, in which
work is transformed into an increase in internal energy, are present. For a process to
be reversible, it must not possess these features. If a process is performed quasistatically, the system passes through states of thermodynamic equilibrium, which
may be traversed as well in one direction as in the opposite direction. If there are no
dissipative effects, all the work done by the system during the performance of a
process in one direction can be returned to the system during the reverse process.
A process will be reversible when it is performed in such a way that the system
is at all times infinitesimally near a state of thermodynamic equilibrium and in the
absence of dissipative effect of any form. Reversible processes are, therefore, purely
ideal, limiting cases of actual processes.
131
Second Law of Thermodynamics
6.11 CARNOT CYCLE
A reversible cycle is an ideal hypothetical cycle in which all the processes constituting the cycle are reversible. Carnot cycle is a reversible cycle. For a stationary
system, as in a piston and cylinder machine, the cycle consists of the following four
successive processes (Fig. 6.20):
(a) A reversible isothermal process in which heat Q1 enters the system at t1 reversibly from a constant temperature source at t1 when the cylinder cover is in
contact with the diathermic cover A. The internal energy of the system increases.
From First law,
Q1 = U2 – U1 + W1–2
(6.14)
(for an ideal gas only, U1 = U2 )
Source, t1
Diathermic cover (A)
Q1
WE
Adiabatic cover (B)
WP
Q2
System
Adiabatic
Sink, t2
Fig. 6.20
Carnot Heat Engine—Stationary System
(b) A reversible adiabatic process in which the diathermic cover A is replaced
by the adiabatic cover B, and work WE is done by the system adiabatically and
reversibly at the expense of its internal energy, and the temperature of the system
decreases from t1 to t2.
Using the first law,
0 = U3 – U2 + W2 – 3
(6.15)
(c) A reversible isothermal process in which B is replaced by A and heat Q2
leaves the system at t 2 to a constant temperature sink at t 2 reversibly, and the internal energy of the system further decreases.
From the first law,
– Q2 = U4 – U3 – W3 – 4
(6.16)
only for an ideal gas,
U3 = U4
(d) A reversible adiabatic process in which B again replaces A, and work Wp is
done upon the system reversibly and adiabatically, and the internal energy of the
system increases and the temperature rises from t2 to t1.
132
Engineering Thermodynamics
Applying the first law,
0 = U1 – U4 – W4 – 1
(6.17)
Two reversible isotherms and two reversible adiabatics constitute a Carnot cycle,
which is represented in p-v coordinates in Fig. 6.21.
Rev. adiabatics
p
1
Q1
WP
2
Rev. isotherm (t1)
4
WE
Rev. isotherm (t2)
Q2
3
v
Carnot Cycle
Fig. 6.21
Summing up Eqs. (6.14) to (6.17),
Q1 – Q2 = (W1 – 2 + W2 – 3) – (W3 – 4 + W4 – 1 )
∑ Q net = ∑ Wnet
or
cycle
cycle
A cyclic heat engine operating on the Carnot cycle is called a Carnot heat engine.
For a steady flow system, the Carnot cycle is represented as shown in Fig. 6.22.
Here heat Q1 is transferred to the system reversibly and isothermally at t1 in the heat
exchanger A, work WT is done by the system reversibly and adiabatically in the
turbine (B), then heat Q2 is transferred from the system reversibly and isothermally
at t2 in the heat exchanger (C), and then work Wp is done upon the system reversibly
and adiabatically by the pump (D). To satisfy the conditions for the Carnot cycle,
there must not be any friction or heat transfer in the pipelines through which the
working fluid flows.
Source, t1
Q1
Flow
Heat exchanger (A)
t1
WP
Pump (D)
t2
System
boundary
t1
t1
WT
Turbine (B)
t2
Heat exchanger (C)
t2
Wnet = Wt - WP
= Q1 - Q 2
Flow
Q2
4
Fig. 6.22
Sink, t2
Carnot Heat Engine—Steady Flow System
133
Second Law of Thermodynamics
6.12 REVERSED HEAT ENGINE
Since all the processes of the Carnot cycle are reversible, it is possible to imagine
that the processes are individually reversed and carried out in reverse order. When
a reversible process is reversed, all the energy transfers associated with the process
are reversed in direction, but remain the same in magnitude. The reversed Carnot
cycle for a steady flow system is shown in Fig. 6.23. The reversible heat engine and
the reversed Carnot heat engine are represented in block diagrams in Fig. 6.24. If E
is a reversible heat engine (Fig. 6.24a), and if it is reversed (Fig. 6.24b), the quantities Q1, Q2 and W remain the same in magnitude, and only their directions are
reversed. The reversed heat engine $ takes heat from a low temperature body, dis
charges heat to a high temperature body, and receives an inward flow of network.
The names heat pump and refrigerator are applied to the reversed heat engine,
which have already been discussed in Sec. 6.6, where the working fluid flows
through the compressor (B), condenser (A), expander (D), and evaporator (C ) to
complete the cycle.
t1
Q1
Flow
t1
WP
t1
Heat exchanger (A)
t1
System
boundry
Pump
(D)
Turbine
(B)
t2
t2
Heat exchanger (C)
Flow
WT
t2
Q2
t2
Fig. 6.23
Reversed Carnot Heat Engine—Steady Flow Process
t1
t1
Q1
Q1
D
E
C
A
B
WT
WP
Wnet = WT - WP
E
A
WP
D
C
B
WT
Wnet = WT - WP
Q2
Q2
t2
t2
(a)
(b)
Fig. 6.24
Carnot Heat Engine and Reversed Carnot Heat Engine
Shown in Block Diagrams
134
6.13
Engineering Thermodynamics
CARNOT’S THEOREM
It states that of all heat engines operating between a given constant temperature
source and a given constant temperature sink, none has a higher efficiency than a
reversible engine.
Let two heat engines EA and EB operate between the given source at temperature
t1 and the given sink at temperature t 2 as shown in Fig. 6.25.
Source, t1
Q1B
Q1A
EA
WA
Q2A
EB
WB
Q2B
Sink t2
Fig. 6.25
Two Cyclic Heat Engines EA and EB Operating between the Same
Source and Sink, of which EB is Reversible
Let EA be any heat engine and EB be any reversible heat engine. We have to
prove that the efficiency of EB is more than that of EA. Let us assume that this is not
true and hA > hB. Let the rates of working of the engines be such that
Q1A = Q1B = Q1
Since
hA > hB
WA
W
> B
Q1A
Q1 B
\
WA > WB
Now, let EB be reversed. Since EB is a reversible heat engine, the magnitudes of
heat and work transfer quantities will remain the same, but their directions will be
reversed, as shown in Fig. 6.26. Since WA > WB, some part of WA (equal to WB) may
be fed to drive the reversed heat engine $B.
Since Q1A = Q1B = Q1, the heat discharged by $B may be supplied to EA. The source
may, therefore, be eliminated (Fig. 6.27). The net result is that EA and $B together constitute a heat engine which, operating in a cycle, produces net work WA – WB, while
exchanging heat with a single reservoir at t2. This violates the Kelvin-Planck statement
of the second law. Hence the assumption that hA > hB is wrong.
Therefore
hB ≥ hA
135
Second Law of Thermodynamics
Source, t1
WA WB
EA
WA
WB
B
EB
WA – WB
Q2A
Q2B
Q2B
Q2A
Sink, t2
Sink, t2
Fig. 6.26 EB is Reversed
6.14
Q1B = Q1
E
EA
Q1A = Q1
Q1B
Q1A
Fig. 6.27
EA and $B together Violate the
K-P Statement
COROLLARY OF CARNOT’S THEOREM
The efficiency of all reversible heat engines operating between the same temperature levels is the same.
Let both the heat engines EA and EB (Fig. 6.25) be reversible. Let us assume
hA > hB. Similar to the procedure outlined in the preceding article, if EB is reversed to
run, say, as a heat pump using some part of the work output (WA) of engine EA, we see
that the combined system of heat pump EB and engine EA, becomes a PMM2. So hA
cannot be greater than hB. Similarly, if we assume hB > hA and reverse the engine EA, we
observe that hB cannot be greater than hA.
Therefore
hA = hB
Since the efficiencies of all reversible heat engines operating between the same
heat reservoirs are the same, the efficiency of a reversible engine is independent of
the nature or amount of the working substance undergoing the cycle.
6.15
ABSOLUTE THERMODYNAMIC TEMPERATURE SCALE
The efficiency of any heat engine cycle receiving heat Q1 and rejecting heat Q2 is
given by
Q - Q2
Q
Wnet
= 1
=1– 2
(6.18)
Q1
Q1
Q1
By the second law, it is necessary to have a temperature difference (t1 – t2) to obtain
work for any cycle. We know that the efficiency of all heat engines operating between the same temperature levels is the same, and it is independent of the working
substance. Therefore, for a reversible cycle (Carnot cycle), the efficiency will depend solely upon the temperatures t 1 and t2, at which heat is transferred, or
hrev = f (t1, t2 )
(6.19)
where f signifies some function of the temperatures. From Equations (6.18) and
(6.19)
h=
136
Engineering Thermodynamics
1–
Q2
= f (t1, t2)
Q1
In terms of a new function F
Q1
= F(t 1, t2)
(6.20)
Q2
If some functional relationship is assigned between t1, t2 and Q1/Q2, the equation
becomes the definition of a temperature scale.
Let us consider two reversible heat engines, E1 receiving heat from the source
at t1, and rejecting heat at t2 to E2 which, in turn, rejects heat to the sink at t3
(Fig. 6.28).
Heat reservoir, t1
Q1
W1 = Q1 - Q2
E1
t2
Q1
Q2
E3
W3 = Q1 - Q3
Q2
W2 = Q2 - Q3
E2
Q3
Q3
Heat reservoir, t3
Fig. 6.28 Three Carnot Engines
Q1
Q
= F(t 1, t 2); 2 = F(t 2, t3)
Q2
Q3
E1 and E2 together constitute another heat engine E3 operating between t1 and t3.
Now
\
Q1
= F(t 1, t3)
Q3
Now
Q1
Q /Q
= 1 3
Q2
Q2 / Q3
or
Q1
F ( t1 , t3 )
= F (t1, t2) =
F ( t2 , t3 )
Q2
(6.21)
The temperatures t 1, t2 and t3 are arbitrarily chosen. The ratio Q1/Q2 depends
only on t1 and t2, and is independent of t3. So t3 will drop out from the ratio on the
right in equation (6.21). After it has been cancelled, the numerator can be written as
f (t1), and the denominator as f (t2), where f is another unknown function. Thus
137
Second Law of Thermodynamics
f ( t1 )
Q1
= F (t1, t2) =
f (t2 )
Q2
Since f (t) is an arbitrary function, the simplest possible way to define the absolute thermodynamic temperature T is to let f (t) = T, as proposed by Kelvin. Then,
by definition
Q1
T
= 1
(6.22)
Q2
T2
The absolute thermodynamic temperature
scale is also known as the Kelvin scale. Two
T
temperatures on the Kelvin scale bear the
same relationship to each other as do the heats
Q
absorbed and rejected respectively by a
Wnet
Carnot engine operating between two reserE
voirs at these temperatures. The Kelvin temperature scale is, therefore, independent of the
Qt
peculiar characteristics of any particular subTt = 273.16 K
stance.
The heat absorbed Q1 and the heat rejected
Q2 during the two reversible isothermal pro- Fig. 6.29 Carnot Heat Engine with
Sink at Triple Point of
cesses bounded by two reversible adiabatics
Water
in a Carnot engine can be measured. In defining the Kelvin temperature scale also, the triple point of water is taken as the standard reference point. For a Carnot engine operating between reservoirs at temperatures T and Tt, T t being the triple point of water (Fig. 6.29), arbitrarily assigned the
value 273.16 K,
Q
T
=
Qt
Tt
\
T = 273.16
T1
Q
Qt
Q1
(6.23)
If this equation is compared with the
equations given in Article 2.3, it is seen that
in the Kelvin scale, Q plays the role of thermometric property. The amount of heat supply Q changes with change in temperature,
just like the thermal emf in a thermocouple.
That the absolute thermodynamic temperature scale has a definite zero point can
be shown by imagining a series of reversible engines, extending from a source at T1
to lower temperatures (Fig. 6.30).
Since
T1
Q
= 1
T2
Q2
E1
T2
W1 = Q1 - Q2
Q2
Q2
E2
T3
W2 = Q2 - Q3
Q3
Q3
E3
T4
W3 = Q3 - Q4
Q4
Q4
Fig. 6.30
Heat Engines Operating in
Series
138
Engineering Thermodynamics
\
T1 - T2 Q1 - Q2
=
T2
Q2
or
T1 – T2 = (Q1 – Q2)
T2
Q2
T2 – T3 = (Q2 – Q3)
T3
Q3
= (Q2 – Q 3)
T2
Q2
T3 – T4 = (Q3 – Q4)
T2
Q2
Similarly
and so on.
If T1 – T2 = T2 – T3 = T3 – T4 = ..., assuming equal temperature intervals
Q1 – Q2 = Q2 – Q3 = Q3 – Q4 = ...
or
W1 = W2 = W3 = ...
Conversely, by making the work quantities performed by the engines in series equal
(W1 = W2 = W3 = ...), we will get
T1 – T2 = T2 – T3 = T3 – T4 = ...
at equal temperature intervals. A scale having one hundred equal intervals between
the steam point and the ice point could be realized by a series of one hundred Carnot
engines operating as in Fig. 6.30. Such a scale would be independent of the working
substance.
If enough engines are placed in series to make the total work output equal to Q1 ,
then by the first law the heat rejected from the last engine will be zero. By the
second law, however, the operation of a cyclic heat engine with zero heat rejection
cannot be achieved, although it may be approached as a limit. When the heat rejected approaches zero, the temperature of heat rejection also approaches zero as a
limit. Thus it appears that a definite zero point exist on the absolute temperature
scale but this point cannot be reached without a violation of the second law.
Also, since Q2 = 0, the isothermal process at T2 = 0 would also be adiabatic
rendering Carnot cycle ambiguous.
Thus any attainable value of absolute temperature is always greater than zero.
This is also known as the Third Law of Thermodynamics which may be stated as
follows: It is impossible by any procedure, no matter how idealized, to reduce any
system to the absolute zero of temperature in a finite number of operations.
This is what is called the Fowler-Guggenheim statement of the third law. The
third law itself is an independent law of nature, and not an extension of the second
law. The concept of heat engine is not necessary to prove the non-attainability of
absolute zero of temperature by any system in a finite number of operations.
139
Second Law of Thermodynamics
6.16
EFFICIENCY OF THE REVERSIBLE HEAT ENGINE
The efficiency of a reversible heat engine in which heat is received solely at T1 is
found to be
h rev = h max = 1 –
FG Q IJ
HQ K
2
1
or
h rev =
=1–
rev
T2
T1
T1 - T2
T1
It is observed here that as T2 decreases, and T1 increases, the efficiency of the re
versible cycle increases.
Since h is always less than unity, T2 is always greater than zero and positive.
The COP of a refrigerator is given by
(COP)refr =
Q2
1
=
Q1
Q1 - Q2
-1
Q2
For a reversible refrigerator, using
Q1
T
= 1
Q2
T2
[COP refr]rev =
T2
T1 - T2
(6.24)
Similarly, for a reversible heat pump
[COPH.P .]rev =
T1
T1 - T2
(6.25)
6.17 EQUALITY OF IDEAL GAS TEMPERATURE AND KELVIN
TEMPERATURE
Let us consider a Carnot cycle executed by an ideal gas, as shown in Fig. 6.31.
The two isothermal processes a–b and c–d are represented by equilateral hyperbolas whose equations are respectively
pV = nR q 1
and
pV = nR q 2
For any infinitesimal reversible process of an ideal gas, the first law may be written
as
d-Q = Cv dq + pdV
Applying this equation to the isothermal process a–b, the heat absorbed is found to
be
V
V nRq 1
V
Q1 = V b pdV = V b
dV = nRq 1 ln b
a
a
Va
V
z
z
140
Engineering Thermodynamics
Reversible
Adiabatics
a
p
Q1
WC
b
Reversible
Isotherms
q1
d
WE
Q2
c
q2
v
Fig. 6.31
Carnot Cycle of an Ideal Gas
Similarly, for the isothermal process c–d, the heat rejected is
Q2 = nRq2 ln
Vc
Vd
V
θ 1 ln b
Va
\
V
θ 2 ln c
Vd
Since the process b–c is adiabatic, the first law gives
Q1
=
Q2
–Cv dq = pdV =
(6.26)
nRq
dV
V
1 q1
dq
V
Cv
= ln c
q
q
Vb
nR 2
z
Similarly, for the adiabatic process d–a
1 q1
dq
V
Cv
= ln d
q
q
Va
nR 2
z
\
or
ln
Vc
V
= ln d
Vb
Va
Vc
V
= d
Vb
Va
Vb
V
= c
Va
Vd
Equation (6.26) thus reduces to
or
Q1
q
= 1
q2
Q2
(6.27)
(6.28)
Second Law of Thermodynamics
141
Kelvin temperature was defined by Eq. (6.22)
Q1
T
= 1
Q2
T2
If q and T refer to any temperature, and q t and Tt refer to the triple point of water,
q
T
=
qt
Tt
Since q T = T t = 273.16 K, it follows that
q=T
(6.29)
The Kelvin temperature is, therefore, numerically equal to the ideal gas temperature and may be measured by means of a gas thermometer.
6.18 TYPES OF IRREVERSIBILITY
It has been discussed in Sec. 6.9 that a process becomes irreversible if it occurs due
to a finite potential gradient like the gradient in temperature or pressure, or if there
is dissipative effect like friction, in which work is transformed into internal energy
increase of the system. Two types of irreversibility can be distinguished:
(a) Internal irreversibility
(b) External irreversibility
The internal irreversibility is caused by the internal dissipative effects like friction,
turbulence, electrical resistance, magnetic hysteresis, etc. within the system. The
external irreversibility refers to the irreversibility occurring at the system boundary
like heat interaction with the surroundings due to a finite temperature gradient.
Sometimes, it is useful to make other distinctions. If the irreversibility of a process is due to the dissipation of work into the increase in internal energy of a system, or due to a finite pressure gradient, it is called mechanical irreversibility. If the
process occurs on account of a finite temperature gradient, it is thermal irreversibility, and if it is due to a finite concentration gradient or a chemical reaction, it is
called chemical irreversibility.
A heat engine cycle in which there is a temperature difference (i) between the
source and the working fluid during heat supply, and (ii) between the working fluid
and the sink during heat rejection, exhibits external thermal irreversibility. If the
real source and sink are not considered and hypothetical reversible processes for
heat supply and heat rejection are assumed, the cycle can be reversible. With the
inclusion of the actual source and sink, however, the cycle becomes externally irreversible.
Solved Examples
Example 6.1 A cyclic heat engine operates between a source temperature of
800°C and a sink temperature of 30°C. What is the least rate of heat rejection per
kW net output of the engine?
142
Engineering Thermodynamics
Solution For a reversible engine, the rate of heat rejection will be minimum
(Fig. 6.32).
T1 = 1073 K
Source
Q1
W = Q1– Q2 = 1 kW
HE
Q2
Sink
T2 = 303 K
Fig. 6.32
T2
T1
30 + 273
=1–
800 + 273
= 1 – 0.282 = 0.718
h max = h rev = 1 –
Now
Wnet
= h max = 0.718
Q1
1
= 1.392 kW
0.718
Now
Q2 = Q1 – Wnet = 1.392 – 1 = 0.392 kW
This is the least rate of heat rejection.
\
Q1 =
Example 6.2 A domestic food freezer maintains
a temperature of – 15°C. The ambient air temperature is 30°C. If heat leaks into the freezer at the
continuous rate of 1.75 kJ/s what is the least power
necessary to pump this heat out continuously?
Freezer temperature,
T2 = – 15 + 273 = 258 K
Ambient air temperature,
T1 = 30 + 273 = 303 K
The refrigerator cycle removes heat from the
freezer at the same rate at which heat leaks into it
(Fig. 6.33).
For minimum power requirement
Q
Q2
= 1
T2
T1
Ambient air T1= 303 K
Q1
Solution
W
R
Q2
Freezer T2 = 258 K
Q2 = 1.75 kJ/s
Fig. 6.33
Second Law of Thermodynamics
143
1.75
¥ 303 = 2.06 kJ/s
2.8
W = Q1 – Q2
\
Q1 =
\
= 2.06 – 1.75 = 0.31 kJ/s = 0.31 kW
Example 6.3 An ideal gas cycle is represented by a rectangle on a p-V diagram.
If p1 and p2 are the lower and higher pressures; and V1 and V2, the smaller and
larger volumes, respectively, then (a) calculate the work done per cycle, (b) indicate which parts of the cycle involve heat flow into the gas. (c) Show that
g -1
if heat capacities are constant.
g p2
V1
+
p2 - p1 V2 - V1
h=
Solution
p2
b
c
a
d
p
p1
V1
V
V2
Fig. 6.34
(a) W = area of the cycle (Fig. 6.34)
= (p2 – p1) (V2 – V1) Ans.
(b) Processes ab and bc
Heat absorbed by 1 mole of gas in one cycle,
Q = Qab + Qbc = Cv (Tb – Ta) + Cp (Tc – Tb)
Now, Ta = Tb
p1
and p2 V1 = R Tb
p2
Tc = Tb
V2
pV
, Tb = 2 1
V1
R
p ˆ
Ê
ÊV
ˆ
Q = Cv Tb Á1 - 1 ˜ + Cp Tb Á 2 - 1˜
Ë
V1 ¯
p2 ¯
Ë
=
p2V1 Ê
p - p1
V -V ˆ
+ Cp 2 1 ˜
Cp 2
p2
V1 ¯
R ÁË
Ans.
144
(c) Q =
Engineering Thermodynamics
( p2 - p1) (V2 - V1)
W
=
Q
p2V1 È
p - p1
V -V ˘
+ Cv 2 1 ˙
Cp 2
R ÍÎ
p2
V1 ˚
=
=
R ( p2 - p1) (V2 - V1 )
CvV1 ( p2 - p1) + C p p2 (V2 - V1 )
C p - Cv
V1
p2
+ Cp
Cv
p2 - p1
V2 - V1
=
g -1
g p2
V1
+
V2 - V1 p2 - p1
Ans.
Example 6.4 A Carnot engine absorbs 200 J of heat from a reservoir at the
temperature of the normal boiling point of water and rejects heat to a reservoir at
the temperature of the triple point of water. Find the heat rejected, the work done
by the engine and the thermal efficiency.
Solution
Q1 = 200 J at T1 = 373.15 K
T2 = 273.16 K
Q2 = Q1
273.16
T2
= 200 ¥
= 146.4 J Ans.
373.15
T1
W = Q1 – Q2 = 53.6 J.
h=
Ans.
53.6
W
=
= 0.268
200
Q1
Ans.
Example 6.5 A reversible heat engine operates between two reservoirs at temperatures of 600°C and 40°C. The engine drives a reversible refrigerator which
operates between reservoirs at temperatures of 40°C and – 20°C. The heat transfer
to the heat engine is 2000 kJ and the net work output of the combined engine
refrigerator plant is 360 kJ.
(a) Evaluate the heat transfer to the refrigerant and the net heat transfer to the
reservoir at 40°C.
(b) Reconsider (a) given that the efficiency of the heat engine and the COP of the
refrigerator are each 40% of their maximum possible values.
Solution
(a) Maximum efficiency of the heat engine cycle (Fig. 6.35) is given by
313
T
hmax = 1 – 2 = 1 –
= 1 – 0.358 = 0.642
873
T1
Again
W1
= 0.642
Q1
\
W1 = 0.642 ¥ 2000 = 1284 kJ
Second Law of Thermodynamics
T1= 873 K
T3 = 253 K
Q4
Q1 = 2000 kJ
W1
W2
HE
R
Q2
W = 360 kJ
Q3 = Q4 + W2
T2 = 313 K
Fig. 6.35
Maximum COP of the refrigerator cycle
(COP)max =
Also
Since
COP =
253
T3
=
= 4.22
313 - 253
T2 - T3
Q4
= 4.22
W2
W1 – W2 = W = 360 kJ
\
W2 = W1 – W = 1284 – 360 = 924 kJ
\
Q4 = 4.22 ¥ 924 = 3899 kJ
\
Q3 = Q4 + W2 = 924 + 3899 = 4823 kJ
Q2 = Q1 – W1 = 2000 – 1284 = 716 kJ
Heat rejection to the 40°C reservoir
= Q2 + Q3 = 716 + 4823 = 5539 kJ
(b) Efficiency of the actual heat engine cycle
h = 0.4 h max = 0.4 ¥ 0.642
\
W1 = 0.4 ¥ 0.642 ¥ 2000
= 513.6 kJ
\
W2 = 513.6 – 360 = 153.6 kJ
COP of the actual refrigerator cycle
COP =
Q4
= 0.4 ¥ 4.22 = 1.69
W2
Therefore
Q4 = 153.6 ¥ 1.69 = 259.6 kJ
Q3 = 259.6 + 153.6 = 413.2 kJ
Q2 = Q1 – W1 = 2000 – 513.6 = 1486.4 kJ
145
146
Engineering Thermodynamics
Heat rejected to the 40°C reservoir
= Q2 + Q3 = 413.2 + 1486.4 = 1899.6 kJ
Example 6.6 Which is the more effective way to increase the efficiency of a
Carnot engine: to increase T1, keeping T2 constant; or to decrease T2, keeping T1
constant?
Solution
The efficiency of a Carnot engine is given by
T2
T1
h=1–
If T2 is constant
FG ∂h IJ = T
H ∂T K T
1
T2
2
2
1
As T1 increases, h increases, and the slope
FG ∂h IJ decreases (Fig. 6.36). If T is
H ∂T K
1
1
constant,
T2
FG ∂η IJ = - 1
H ∂T K T
2
1
T1
As T2 decreases, h increases, but the slope
FG ∂η IJ remains constant (Fig. 6.37).
H ∂T K
2
T1
1.0
1.0
h
h
Slope = – 1/T
0
0
T1
T2
T1
T2
Fig. 6.36
Also
FG ∂η IJ = T and FG ∂η IJ = - T
H ∂T K T H ∂T K T
1 T
1
Since
Fig. 6.37
2
2
1
T1 > T2,
2
1
2
1
T1
FG ∂η IJ > FG ∂η IJ
H ∂T K H ∂T K
2
T1
1
T2
147
Second Law of Thermodynamics
So, the more effective way to increase the efficiency is to decrease T2. Alternatively, let T2 be decreased by DT with T1 remaining the same
h1 = 1 –
T2 - DT
T1
If T1 is increased by the same DT, T2 remaining the same
h2 = 1 –
T2
T1 + DT
Then
h1 – h2 =
=
T2
T - DT
- 2
T1 + DT
T1
(T1 - T2 ) DT + ( DT )2
T1 ( T1 + DT )
Since
T1 > T2, (h 1 – h 2) > 0
The more effective way to increase the cycle efficiency is to decrease T2.
Example 6.7 Kelvin was the first to point out the thermodynamic wastefulness
of burning fuel for the direct heating of a house. It is much more economical to use
the high temperature heat produced by combustion in a heat engine and then to use
the work so developed to pump heat from outdoors up to the temperature desired in
the house. In Fig. 6.38 a boiler furnishes heat Q1 at the high temperature T1. This
heat is absorbed by a heat engine, which extracts work W and rejects the waste
heat Q2 into the house at T2. Work W is in turn used to operate a mechanical refrigerator or heat pump, which extracts Q3 from outdoors at temperature T3 and reject
Q¢2 (where Q ¢2 = Q 3 + W) into the house. As a result of this cycle of operations, a
total quantity of heat equal to Q2 + Q¢2 is liberated in
Boiler
the house, against Q1 which would be provided diQ1
T1
rectly by the ordinary combustion of the fuel. Thus the
Q1
ratio (Q2 + Q¢2 )/Q1 represents the heat multiplication
factor of this method. Determine this multiplication
CHE
factor if T 1 = 473 K, T2 = 293 K, and T3 = 273 K.
W
Solution
For the reversible heat engine (Fig. 6.38)
Q2
T
Q2
= 2
Q1
T1
\
Q2 = Q1
House
T2
FG T IJ
HT K
Q¢2 = Q3 + W
2
1
Also
T -T
W
h=
= 1 2
Q1
T1
or
T -T
W = 1 2 ◊ Q1
T1
W
H.P.
Q3
Outdoors
Q3
T3
Fig. 6.38
148
Engineering Thermodynamics
For the reversible heat pump
COP =
Q2¢
T2
=
W
T2 - T3
T2
T - T ◊ Q1
◊ 1 2
T2 - T3
T1
\
Q¢2 =
\
Multiplication factor (M.F.)
Q + Q2¢
= 2
=
Q1
Q1
T2
T2
T -T
+ Q1 ◊
◊ 1 2
T1
T2 - T3
T1
Q1
or
M.F. =
T22 - T2 T3 + T2 T1 - T22
T1 ( T2 - T3 )
or
M.F. =
T2 ( T1 - T3 )
T1 ( T2 - T3 )
Here
\
T1 = 473 K, T2 = 293 K and T3 = 273 K
M.F. =
293( 473 - 273) 2930
=
= 6.3
473
473 ( 293 - 273)
which means that every kg of coal burned would deliver the heat equivalent to over
6 kg. Of course, in an actual case, the efficiencies would be less than Carnot efficiencies, but even with a reduction of 50%, the possible savings would be quite
significant.
Example 6.8 It is proposed that solar energy be used to warm a large collector
plate. This energy would, in turn, be transferred as heat to a fluid within a heat
engine, and the engine would reject energy as heat to the atmosphere. Experiments
indicate that about 1880 kJ/m2 h of energy can be collected when the plate is operating at 90°C. Estimate the minimum collector area that would be required for a
plant producing 1 kW of useful shaft power. The atmospheric temperature may be
assumed to be 20°C.
Solution The maximum efficiency for the heat engine operating between the collector plate temperature and the atmospheric temperature (Fig. 6.39) as follows:
293
T
h max = 1 - 2 = 1 = 0.192
363
T1
The efficiency of any actual heat engine operating between these temperatures
would be less than this efficiency.
\
Q min =
1 kJ/s
W
=
= 5.21 kJ/s
0.192
h max
= 18,800 kJ/h
149
Second Law of Thermodynamics
\
Minimum area required for the collector plate
=
18,800
= 10 m2
1880
Example 6.9 A reversible heat engine in a satellite operates between a hot reservoir at T1 and a radiating panel at T2. Radiation from the panel is proportional
to its area and to T 24. For a given work output and value of T1 show that the area of
the panel will be minimum when
T2
= 0.75.
T1
Determine the minimum area of the panel for an output
of 1 kW if the constant of proportionality is 5.67 ¥ 10–8 W/
m2 K4 and T1 is 1000 K.
Solution For the heat engine (Fig. 6.39), the heat rejected
Q2 to the panel (at T2) is equal to the energy emitted from
the panel to the surroundings by radiation. If A is the area
of the panel, Q2 µ AT 42, or Q2 = KAT42 , where K is a constant.
Now,
T -T
W
h=
= 1 2
Q1
T1
Q1
HE
W
Q2 = KAT 24
Panel
T2
Q2
Q
Q
KAT24
W
= 1= 2 =
T1 - T2
T1 T2
T2
or
T1
Fig. 6.39
= KAT 32
\
A=
W
KT23 ( T1 - T2 )
=
W
K ( T1T23 - T24 )
For a given W and T1, A will be minimum when
W
dA
=(3T1T 22 – 4T 32) ◊ (T1T 32 – T 42) – 2 = 0
K
dT2
Since
(T1T 32 – T 42) –2 π 0, 3T1T 22 = 4T 32
\
T2
= 0.75 Proved.
T1
∵
Amin =
=
Here
W
3
K ( 0.75) T13 ( T1 - 0.75T1 )
256 W
W
=
27 4
27 KT14
K
T1
256
W = 1 kW, K = 5.67 ¥ 10–8 W/m2 K4, and T1 = 1000 K
150
\
Engineering Thermodynamics
Amin =
=
256 ¥ 1kW ¥ m 2 K 4
27 ¥ 5.67 ¥ 10 -8 W ¥ (1000 ) 4 K 4
256 ¥ 10 3
m2 = 0.1672 m2
27 ¥ 5.67 ¥ 10 -8 ¥ 1012
Summary
A heat engine cycle is a thermodynamic cycle in which there is a net flow of heat
to the system and a net flow of work from the system. The system which executes
a heat engine cycle is a heat engine. The efficiency of a heat engine, or of its
cycle is
h=
Wnet
Q1
where Q1 is the heat transferred to the system in a cycle, and Wnet is the net work
of the cycle. This efficiency is called thermal efficiency.
A heat pump or a refrigerator is a system to which there is a net flow of work
and from which there is a net flow of heat in a cycle.
The second law is stated as follows (after Kelvin and Planck): It is impossible
for a heat engine to produce net work in a complete cycle if it exchanges heat
only with bodies at a single fixed temperature.
An equivalent statement (after Clausius) follows: It is impossible for a system
working in a complete cycle to accomplish as its sole effect the transfer of heat
from a body at a given temperature to a body at a higher temperature.
A process is reversible if, after the process has been carried out, it is possible
by any means whatsoever to restore both the system and the entire surroundings
to exactly the same states that were in before the process. A process that is not
reversible is irreversible. A reversible process is a process that can be undone in
such a way that no trace remains anywhere of the fact that the process had
occurred. The causes of irreversibility of a process are:
1. Finite potential gradient causing the process like temperature, pressure, concentration, etc.
2. Presence of dissipative effects like friction in which macroscopic work dissipates into an increase of internal energy and then heat.
A reversible process would require infinite time.
A reversible cycle is a cycle composed entirely of reversible processes. The
classical example is the Carnot cycle which consists of two reversible isothermal
processes and two reversible adiabatic processes.
No heat engine operating between fixed temperature levels can be more
efficient than a reversible engine operating between the same temperatures. The
efficiency of all reversible heat engines operating between the same temperature
Second Law of Thermodynamics
151
levels is the same. The efficiency of a reversible heat engine is independent of
the nature or amount of the working substance undergoing the cycle. The absolute
thermodynamic scale is defined by the relation
T1 Q1
=
T2 Q2
where Q1 is the heat received from a source at T1 and Q2 is the heat rejected to a
sink at T2 by a reversible heat engine. The efficiency of a reversible heat engine
receiving heat solely at T1 and rejecting heat solely at T2 is given by
hmax = hrev =
T1 - T2
T
=1– 2
T1
T1
The COPs of a Carnot refrigerator and a Carnot heat pump are
(COPmax)Ref =
T2
T1
, (COPmax)HP =
T1 - T2
T1 - T2
The absolute thermodynamic temperature scale or the Kelvin scale is equivalent
to the ideal gas temperature scale, and the Kelvin temperature can be measured
by a gas thermometer.
Review Questions
6.1
What is the qualitative difference between heat and work? Why are heat and
work not completely interchangeable forms of energy?
6.2 What is a cyclic heat engine?
6.3 Explain a heat engine cycle performed by a closed system.
6.4 Explain a heat engine cycle performed by a steady flow system.
6.5 Define the thermal efficiency of a heat engine cycle. Can this be 100%?
6.6 Draw a block diagram showing the four energy interactions of a cyclic heat engine.
6.7 What is a thermal energy reservoir? Explain the terms ‘source’ and ‘sink’.
6.8 What is a mechanical energy reservoir?
6.9 Why can all processes in a TER or an MER be assumed to be quasi-static?
6.10 Give the Kelvin-Planck statement of the second law.
6.11 To produce net work in a thermodynamic cycle, a heat engine has to exchange
heat with two thermal reservoirs. Explain.
6.12 What is a PMM2? Why is it impossible?
6.13 Give the Clausius’ statement of the second law.
6.14 Explain the operation of a cyclic refrigerator plant with a block diagram.
6.15 Define the COP of a refrigerator.
6.16 What is a heat pump? How does it differ from a refrigerator?
6.17 Can you use the same plant as a heat pump in winter and as a refrigerator in
summer? Explain.
6.18 Show that the COP of a heat pump is greater than the COP of a refrigerator by
unity.
152
Engineering Thermodynamics
6.19 Why is direct heating thermodynamically wasteful?
6.20 How can a heat pump upgrade low grade waste heat?
6.21 Establish the equivalence of Kelvin-Planck and Clausius statements.
6.22 What is a reversible process? A reversible process should not leave any evidence
to show that the process had ever occurred. Explain.
6.23 How is a reversible process only a limiting process, never to be attained in practice?
6.24 All spontaneous processes are irreversible. Explain.
6.25 What are the causes of irreversibility of a process?
6.26 Show that heat transfer through a finite temperature difference is irreversible.
6.27 Demonstrate, using the second law, that free expansion is irreversible.
6.28 What do you understand by dissipative effects? When is work said to be dissipated?
6.29 Explain perpetual motion of the third kind.
6.30 Demonstrate using the second law how friction makes a process irreversible.
6.31 When a rotating wheel is brought to rest by applying a brake, show that the
molecular internal energy of the system (of the brake and the wheel) increases.
6.32 Show that the dissipation of stirring work to internal energy is irreversible.
6.33 Show by second law that the dissipation of electrical work into internal energy or
heat is irreversible.
6.34 What is a Carnot cycle? What are the four processes which constitute the cycle?
6.35 Explain the Carnot heat engine cycle executed by: (a) a stationary system, and
(b) a steady flow system.
6.36 What is a reversed heat engine?
6.37 Show that the efficiency of a reversible engine operating between two given
constant temperatures is the maximum.
6.38 Show that the efficiency of all reversible heat engines operating between the
same temperature levels is the same.
6.39 Show that the efficiency of a reversible engine is independent of the nature or
amount of the working substance going through the cycle.
6.40 How does the efficiency of a reversible cycle depend only on the two temperatures at which heat is transferred?
6.41 What is the absolute thermodynamic temperature scale? Why is it called absolute?
6.42 How is the absolute scale independent of the working substance?
6.43 How does Q play the role of thermometric property in the Kelvin scale?
6.44 Show that a definite zero point exists on the absolute temperature scale but that
this point cannot be reached without a violation of the second law.
6.45 Give the Fowler-Guggenheim statement of the third law.
6.46 Is the third law an extension of the second law? Is it an independent law of
nature? Explain.
6.47 How does the efficiency of a reversible engine vary as the source and sink temperatures are varied? When does the efficiency become 100%?
6.48 For a given T2, show that the COP of a refrigerator increases as T1 decreases.
6.49 Explain how the Kelvin temperature can be measured with a gas thermometer.
6.50 Establish the equality of ideal gas temperature and Kelvin temperature.
6.51 What do you understand by internal irreversibility and external irreversibility?
Second Law of Thermodynamics
153
6.52 Explain mechanical, thermal and chemical irreversibilities.
6.53 A Carnot engine with a fuel burning device as source and a heat sink cannot be
treated as a reversible plant. Explain.
Problems
6.1 An inventor claims to have developed an engine that takes in 105 MJ at a temperature of 400 K, rejects 42 MJ at a temperature of 200 K, and delivers 15 kWh
of mechanical work. Would you advise investing money to put this engine in the
market?
Ans. No
6.2 If a refrigerator is used for heating purposes in winter so that the atmosphere
becomes the cold body and the room to be heated becomes the hot body, how
much heat would be available for heating for each kW input to the driving
motor? The COP of the refrigerator is 5, and the electromechanical efficiency of
the motor is 90%. How does this compare with resistance heating?
Ans. 5.4 kW
6.3 Using an engine of 30% thermal efficiency to drive a refrigerator having a COP
of 5, what is the heat input into the engine for each MJ removed from the cold
body by the refrigerator?
Ans. 666.67 kJ
If this system is used as a heat pump, how many MJ of heat would be available
for heating for each MJ of heat input to the engine?
Ans. 1.8 MJ
6.4 An electric storage battery which can exchange heat only with a constant temperature atmosphere goes through a complete cycle of two processes. In process
1–2, 2.8 kWh of electrical work flow into the battery while 732 kJ of heat flow
out to the atmosphere. During process 2–1, 2.4 kWh of work flow out of the
battery. (a) Find the heat transfer in process 2–1. (b) If the process 1–2 has occurred as above, does the first law or the second law limit the maximum
possible work of process 2–1? What is the maximum possible work? (c) If the
maximum possible work were obtained in process 2–1, what will be the heat
transfer in the process?
Ans (a) – 708 kJ (b) Second law, W2 – 1 = 9348 kJ (c) Q2 – 1 = 0
6.5 A household refrigerator is maintained at a temperature of 2°C. Every time the
door is opened, warm material is placed inside, introducing an average of
420 kJ, but making only a small change in the temperature of the refrigerator.
The door is opened 20 times a day, and the refrigerator operates at 15% of the
ideal COP. The cost of work is Rs. 2.50 per kWh. What is the monthly bill for
this refrigerator? The atmosphere is at 30°C.
Ans Rs. 118.80
6.6 A heat pump working on the Carnot cycle takes in heat from a reservoir at 5°C
and delivers heat to a reservoir at 60°C. The heat pump is driven by a reversible
heat engine which takes in heat from a reservoir at 840°C and rejects heat to a
reservoir at 60°C. The reversible heat engine also drives a machine that absorbs
30 kW. If the heat pump extracts 17 kJ/s from the 5°C reservoir, determine (a) the
rate of heat supply from the 840°C source, and (b) the rate of heat rejection to the
60°C sink.
Ans. (a) 47.61 kW; (b) 34.61 kW
6.7 A refrigeration plant for a food store operates as a reversed Carnot heat engine
cycle. The store is to be maintained at a temperature of – 5°C and the heat transfer from the store to the cycle is at the rate of 5 kW. If heat is transferred from the
154
Engineering Thermodynamics
cycle to the atmosphere at a temperature of 25°C, calculate the power required to
drive the plant.
Ans. 0.56 kW
6.8 A heat engine is used to drive a heat pump. The heat transfers from the heat
engine and from the heat pump are used to heat the water circulating through the
radiators of a building. The efficiency of the heat engine is 27% and the COP of
the heat pump is 4. Evaluate the ratio of the heat transfer to the circulating water
to the heat transfer to the heat engine.
Ans. 1.81
6.9 If 20 kJ are added to a Carnot cycle at a temperature of 100°C and 14.6 kJ are
rejected at 0°C, determine the location of absolute zero on the Celsius scale.
Ans. – 270.37°C
6.10 Two reversible heat engines A and B are arranged in series, A rejecting heat
directly to B. Engine A receives 200 kJ at a temperature of 421°C from a hot
source, while engine B is in communication with a cold sink at a temperature of
4.4°C. If the work output of A is twice that of B, find (a) the intermediate temperature between A and B, (b) the efficiency of each engine, and (c) the heat
rejected to the cold sink.
Ans. 143.4°C, 40% and 33.5%, 80 kJ
6.11 A heat engine operates between the maximum and minimum temperatures of
671°C and 60°C respectively, with an efficiency of 50% of the appropriate Carnot
efficiency. It drives a heat pump which uses river water at 4.4°C to heat a block
of flats in which the temperature is to be maintained at 21.1°C. Assuming that a
temperature difference of 11.1°C exists between the working fluid and the river
water, on the one hand, and the required room temperature on the other, and
assuming the heat pump to operate on the reversed Carnot cycle, but with a COP
of 50% of the ideal COP, find the heat input to the engine per unit heat output
from the heat pump. Why is direct heating thermodynamically more wasteful?
Ans. 0.79 kJ/kJ heat input
6.12 An ice-making plant produces ice at atmospheric pressure and at 0°C from
water. The mean temperature of the cooling water circulating through the
condenser of the refrigerating machine is 18°C. Evaluate the minimum electrical
work in kWh required to produce 1 tonne of ice. (The enthalpy of fusion of ice at
atmospheric pressure is 333.5 kJ/kg).
Ans. 6.11 kWh
6.13 A reversible engine works between three thermal reservoirs, A, B and C. The
engine absorbs an equal amount of heat from the thermal reservoirs A and B kept
at temperatures TA and TB respectively, and rejects heat to the thermal
reservoir C kept at temperature TC. The efficiency of the engine is a times the
efficiency of the reversible engine, which works between the two reservoirs A
and C. Prove that
TA
T
= (2a – 1) + 2 (1 – a) A
TB
TC
6.14 A reversible engine operates between temperatures T1 and T (T1 > T). The
energy rejected from this engine is received by a second reversible engine
at the same temperature T. The second engine rejects energy at temperature
T2 (T2 < T1). Show that (a) temperature T is the arithmetic mean of temperatures
T1 and T2 if the engines produce the same amount of work output, and (b) temperature T is the geometric mean of temperatures T1 and T2 if the engines have
the same cycle efficiencies.
Second Law of Thermodynamics
155
6.15 Two Carnot engines A and B are connected in series between two thermal reservoirs maintained at 1000 K and 100 K respectively. Engine A receives 1680 kJ of
heat from the high-temperature reservoir and rejects heat to the Carnot engine B.
Engine B takes in heat rejected by engine A and rejects heat to the low-temperature reservoir. If engines A and B have equal thermal efficiencies, determine (a)
the heat rejected by engine B, (b) the temperature at which heat is rejected by
engine, A, and (c) the work done during the process by engines, A and B respectively. If engines A and B deliver equal work, determine (d) the amount of heat
taken in by engine B, and (e) the efficiencies of engines A and B.
Ans. (a) 168 kJ, b) 316.2 K, (c) 1148.7, 363.3 kJ, (d) 924 kJ, (e) 45%, 81.8%.
6.16 A heat pump is to be used to heat a house in winter and then reversed to cool the
house in summer. The interior temperature is to be maintained at 20°C. Heat
transfer through the walls and roof is estimated to be 0.525 kJ/s per degree temperature difference between the inside and outside. (a) If the outside temperature
in winter is 5°C, what is the minimum power required to drive the heat pump?
(b) If the power output is the same as in part (a), what is the maximum outer
temperature for which the inside can be maintained at 20°C?
Ans. (a) 403 W, (b) 35.4°C.
6.17 Consider an engine in outer space which operates on the Carnot cycle. The only
way in which heat can be transferred from the engine is by radiation. The rate at
which heat is radiated is proportional to the fourth power of the absolute
temperature and to the area of the radiating surface. Show that for a given power
output and a given T1, the area of the radiator will be a minimum when
T2
3
=
T1
4
6.18 It takes 10 kW to keep the interior of a certain house at 20°C when the outside
temperature is 0°C. This heat flow is usually obtained directly by burning gas or
oil. Calculate the power required if the 10 kW heat flow were supplied by operating a reversible engine with the house as the upper reservoir and the outside
surroundings as the lower reservoir, so that the power were used only to perform
work needed to operate the engine.
Ans. 0.683 kW
6.19 Prove that the COP of a reversible refrigerator operating between two given temperatures is the maximum.
6.20 A house is to be maintained at a temperature of 20°C by means of a heat pump
pumping heat from the atmosphere. Heat losses through the walls of the house
are estimated at 0.65 kW per unit of temperature difference between the inside of
the house and the atmosphere. (a) If the atmospheric temperature is – 10°C, what
is the minimum power required to drive the pump? (b) It is proposed to use the
same heat pump to cool the house in summer. For the same room temperature,
the same heat loss rate, and the same power input to the pump, what is the maximum permissible atmospheric temperature?
Ans. 2 kW, 50°C.
6.21 A solar-powered heat pump receives heat from a solar collector at Th, rejects
heat to the atmosphere at Ta, and pumps heat from a cold space at Tc. The three
heat transfer rates are Qh, Qa, and Qc respectively. Derive an expression for the
minimum ratio Qh /Qc, in terms of the three temperatures.
If Th = 400 K, Ta = 300 K, Tc = 200 K, Qc = 12 kW, what is the minimum Qh?
If the collector captures 0.2 kW/m2, what is the minimum collector area required?
Ans. 26.25 kW, 131.25 m2
156
Engineering Thermodynamics
6.22 A heat engine operating between two reservoirs at 1000 K and 300 K is used to
drive a heat pump which extracts heat from the reservoir at 300 K at a rate twice
that at which the engine rejects heat to it. If the efficiency of the engine is 40% of
the maximum possible and the COP of the heat pump is 50% of the maximum
possible, what is the temperature of the reservoir to which the heat pump rejects
heat? What is the rate of heat rejection from the heat pump if the rate of heat
supply to the engine is 50 kW?
Ans. 326.5 K, 86 kW
6.23 A reversible power cycle is used to drive a reversible heat pump cycle. The power
cycle takes in Q1 heat units at T1 and rejects Q2 at T2. The heat pump abstracts Q4
from the sink at T4 and discharges Q3 at T3. Develop an expression for the ratio
Q4/Q1 in terms of the four temperatures.
Q
T (T - T )
Ans. 4 = 4 1 2
Q1
T1 (T3 - T4 )
6.24 Prove that the following propositions are logically equivalent: (a) A PMM2 is
impossible, (b) A weight sliding at constant velocity down a frictional inclined
plane executes an irreversible process.
6.25 A heat engine receives half of its heat supply at 1000 K and half at 500 K while
rejecting heat to a sink at 300 K. What is the maximum thermal efficiency of the
heat engine?
Ans. 55%
6.26 A heat pump provides 3 ¥ 104 kJ/h to maintain a dwelling at 23°C on a day when
the outside temperature is 0°C. The power input to the heat pump is 4 kW. Determine the COP of the heat pump and compare it with the COP of a reversible heat
pump operating between the reservoirs at the same two temperatures.
Ans. 2.08, 12.87
6.27 When the outside temperature is – 10°C, a residential heat pump must provide
3.5 ¥ 106 kJ per day to a dwelling to maintain its temperature at 20°C. If the
electricity costs Rs 2.10 per kWh, find the minimum operating cost for each day
of operation.
Ans. Rs 208.83
6.28 A reversible power cycle receives energy QH from a reservoir at temperature TH
and rejects QC to a reservoir at temperature TC. The work developed by the power
cycle is used to drive a reversible heat pump that removes Q¢C from a reservoir at
T¢C and rejects energy Q¢H to a reservoir at temperature T¢H. (a) Develop an expression for the ratio Q¢H/QH in terms of the temperatures of the four reservoirs. (b)
QH¢
What must be the relationship of the temperatures TH, TC, T¢C and T¢H for Q to
H
exceed a value of unity?
QH¢
TH¢ (TH - TC )
TC TH
<
Ans. (a) Q = T T
,
(b)
TC TH¢
¢ - TC¢ )
H
H( H
7
Entropy
7.1 INTRODUCTION
The first law of thermodynamics was stated in terms of cycles first and it was shown
that the cyclic integral of heat is equal to the cyclic integral of work. When the first
law was applied for thermodynamic processes, the existence of a property, the internal energy, was found. Similarly, the second law was also first stated in terms of
cycles executed by systems. When applied to processes, the second law also leads
to the definition of a new property, known as entropy. If the first law is said to be the
law of internal energy, then second law may be stated to be the law of entropy. In
fact, thermodynamics is the study of three E’s, namely, energy, equilibrium and
entropy.
7.2 TWO REVERSIBLE ADIABATIC PATHS CANNOT
INTERSECT EACH OTHER
P
Let it be assumed that two reversible adiabatics AC and BC intersect each other at
point C (Fig. 7.1). Let a reversible isotherm AB be drawn in such a way that it
intersects the reversible adiabatics at A
and B. The three reversible processes AB,
BC, and CA together constitute a reversA
Rev.
ible cycle, and the area included repreisotherm
B
sents the net work output in a cycle. But
Rev.
such a cycle is impossible, since net work
adiabatics
is being produced in a cycle by a heat engine by exchanging heat with a single resC
ervoir in the process AB, which violates
the Kelvin-Planck statement of the second
v
law. Therefore, the assumption of the intersection of the reversible adiabatics is Fig. 7.1 Assumption of Two Reversible
wrong. Through one point, there can pass
Adiabatics Intersecting Each
only one reversible adiabatic.
Other
158
Engineering Thermodynamics
Since two constant property lines can never intersect each other, it is inferred
that a reversible adiabatic path must represent some property, which is yet to be
identified.
7.3
CLAUSIUS’ THEOREM
P
Let a system be taken from an equilibrium state i to another equilibrium state f by
following the reversible path i-f (Fig. 7.2). Let a reversible adiabatic i-a be drawn
through i and another reversible adiabatic
b-f be drawn through f. Then a reversible
Rev. adiabatics
isotherm a-b is drawn in such a way that
the area under i-a-b-f is equal to the area
i
under i-f. Applying the first law for
b
a
Process i-f
Qi - f = Uf – Ui + Wif
(7.1)
Rev.
isotherm
Process i-a-b-f
Qiabf = Uf – Ui + Wiabf
(7.2)
Since
v
Fig. 7.2
Wif = Wiabf
\
f
From Eqs. (7.1) and (7.2)
Reversible Path Substituted
by Two Reversible Adiabatics
and a Reversible Isotherm
Qif = Qiabf
= Qia + Qab + Qbf
Since
Qia = 0 and Qbf = 0
Qif = Qab
Heat transferred in the process i-f is equal to the heat transferred in the isothermal
process a-b.
Thus any reversible path may be substituted by a reversible zigzag path, between the same end states, consisting of a reversible adiabatic followed by a reversible isotherm and then by a reversible adiabatic, such that the heat transferred during the isothermal process is the same as that transferred during the original process.
Let a smooth closed curve representing a reversible cycle (Fig. 7.3) be considered. Let the closed cycle be divided into a large number of strips by means of
reversible adiabatics. Each strip may be closed at the top and bottom by reversible
isotherms. The original closed cycle is thus replaced by a zigzag closed path consisting of alternate adiabatic and isothermal processes, such that the heat transferred during all the isothermal processes is equal to the heat transferred in the
original cycle. Thus the original cycle is replaced by a large number of Carnot
cycles. If the adiabatics are close to one another and the number of Carnot cycles is
large, the saw-toothed zig-zag line will coincide with the original cycle.
159
Entropy
Rev. adiabatics
P
dQ3
T3 e
Rev. isotherms
f
T1
a
b
T1
dQ1
dQ2
T2
dQ4
c
d
Original reversible
circle
g
T4
h
v
Fig. 7.3
A Reversible Cycle Split into a Large Number of Carnot Cycles
For the elemental cycle abcd d-Q1 heat is absorbed reversibly at T1, and d-Q2
heat is rejected reversibly at T2
dQ
1
T1
=
dQ
2
T2
If heat supplied is taken as positive and heat rejected as negative
dQ
1
+
dQ
2
T1
T2
Similarly, for the elemental cycle efgh
dQ
3
+
=0
dQ
4
=0
T3
T4
If similar equations are written for all the elemental Carnot cycles, then for the
whole original cycle
d-Q1 d-Q 2 d-Q 3 d-Q 4
+
+
+
+º = 0
T1
T2
T3
T4
z
d-Q
=0
(7.3)
R T
The cyclic integral of d-Q/T for a reversible cycle is equal to zero. This is known as
Clausius’ theorem. The letter R emphasizes the fact that the equation is valid only
for a reversible cycle.
or
7.4
THE PROPERTY OF ENTROPY
Let a system be taken from an initial equilibrium state i to a final equilibrium state
f by following the reversible path R1 (Fig. 7.4). The system is brought back from f
to i by following another reversible path R2. Then the two paths R 1 and R2 together
constitute a reversible cycle. From Clausius’ theorem
dQ
=0
R 1R 2 T
z
160
Engineering Thermodynamics
z
z
R1
R2
-
P
The above integral may be replaced as the
sum of two integrals, one for path R1 and
the other for path R2
f dQ
i dQ
+ f
=0
i
T
T
i
R2
R1
-
z dTQ = – z dQT
f
i
R1
or
i
f
f
R2
v
Fig. 7.4
Since R2 is a reversible path
-
-
z dTQ = z dQT
f
i
R1
f
i
Two Reversible Paths R1 and
R2 Between Two Equilibrium
States i and f
R2
Since R1 and R2 represent any two reversible paths,
z
f
dQ
i
T
is independent of
R
the reversible path connecting i and f. Therefore, there exists a property of a
system whose value at the final state f minus its value at the initial state i is equal to
-
z dQT . This property is called entropy, and is denoted by S. If S is the entropy at
f
i
i
R
the initial state i, and Sf is the entropy at the final state f, then
-
z dTQ = S – S
f
f
i
i
(7.4)
R
When the two equilibrium states are infinitesimally near
dQ
R
= dS
(7.5)
T
where dS is an exact differential because S is a point function and a property. The
subscript R in d-Q indicates that heat d-Q is transferred reversibly.
The word ‘entropy’ was first used by Clausius, taken from the Greek word
‘tropee’ meaning ‘transformation’. It is an extensive property, and has the unit
J/K. The specific entropy
S
s=
J/kg K
m
If the system is taken from an initial equilibrium state i to a final equilibrium
state f by an irreversible path, since entropy is a point or state function, and the
entropy change is independent of the path followed, the non-reversible path is to be
replaced by a reversible path to integrate for the evaluation of entropy change in the
irreversible process (Fig. 7.5).
z
f d-Q
rev
= (D S ) irrev path
T
Integration can be performed only on a reversible path.
Sf – S i =
i
(7.6)
161
Entropy
Integration can be Done Only on a Reversible Path
Fig. 7.5
7.5
TEMPERATURE-ENTROPY PLOT
The infinitesimal change in entropy dS due to reversible heat transfer d-Q at temperature T is
d-Q rev
T
If d Qrev = 0, i.e., the process is reversible and adiabatic
dS = 0
and
S = constant
A reversible adiabatic process is, therefore, an isentropic process.
dS =
Now
d- Qrev = TdS
or
z
Qrev =
z
f
TdS
i
The system is taken from i to f reversibly (Fig. 7.6). The area under the curve
f
T dS is equal to the heat transferred in the process.
i
For reversible isothermal heat transfer (Fig. 7.7), T = constant.
T
i
f
i
T
T
R
T
f
f
Qrev = Ú TdS
Q
i
S
dS
Sf
S
Fig. 7.6
Area Under a Reversible
Path on the T-s Plot
Represents Heat Transfer
Si
S
Sf
Fig. 7.7 Reversible Isothermal
Heat Transfer
162
Engineering Thermodynamics
\
Qrev = T
z
i
f
dS = T (Sf – Si)
For a reversible adiabatic process, dS = 0, S = C (Fig. 7.8).
The Carnot cycle comprising two reversible isotherms and two reversible
adiabatics forms a rectangle in the T-S plane (Fig. 7.9). Process 4 – 1 represents
reversible isothermal heat addition Q1 to the system at T1 from an external source,
process 1–2 is the reversible adiabatic expansion of the system producing WE
amount of work, process 2 – 3 is the reversible isothermal heat rejection from the
system to an external sink at T2, and process 3 – 4 represents reversible adiabatic
compression of the system consuming Wc amount of work. Area 1234 represents
the net work output per cycle and the area under 4 – 1 indicates the quantity of heat
added to the system Q1.
Q1
4
1
T1
T
T
i
WE
WC
3
f
Q2
S
Fig. 7.8
\
7.6
T2
S
Reversible Adiabatic
is Isentropic
h Carnot =
=
and
2
Fig. 7.9
Q1 - Q2
Q1
T1 - T2
T1
=
Carnot Cycle
T1 ( S1 - S 4 ) - T2 ( S 2 - S 3 )
T1 ( S1 - S 4 )
=1–
T2
T1
Wnet = Q1 – Q2 = (T1 – T2) (S1 – S4)
THE INEQUALITY OF CLAUSIUS
Let us consider a cycle ABCD (Fig. 7.10). Let AB be a general process, either reversible or irreversible, while the other processes in the cycle are reversible. Let the
cycle be divided into a number of elementary cycles, as shown. For one of these
elementary cycles
dQ
2
h =1– dQ
where d-Q is the heat supplied at T, and d-Q2 the heat rejected at T2 .
Now, the efficiency of a general cycle will be equal to or less than the efficiency
of a reversible cycle.
\
1–
dQ
2
-
dQ
F
GH
£ 1-
I
dQ JK
dQ
2
-
rev
Entropy
Inequality of Clausius
Fig. 7.10
F dQ I
dQ GH dQ JK
dQ F dQ I
£G
dQ
H dQ JK
F dQ I = T
GH dQ JK T
dQ
2
or
-
-
2
≥
-
-
or
163
rev
-
-
-
2
2
rev
-
Since
-
2
-
dQ
\
-
dQ 2
-
or
dQ
£
2
rev
T
T2
-
£
dQ 2
, for any process AB, reversible
T
T2
or irreversible.
For a reversible process
ds =
dQ
rev
T
=
dQ
2
T2
(7.7)
Hence, for any process AB
dQ
T
£ ds
(7.8)
Then for any cycle
z
z
dQ
z
£ ds
T
Since entropy is a property and the cyclic integral of any property is zero
dQ
£0
(7.9)
T
This equation is known as the inequality of Clausius. It provides the criterion of
the reversibility of a cycle.
164
Engineering Thermodynamics
z
z
z
If
dQ
T
dQ
T
= 0, the cycle is reversible,
< 0, the cycle is irreversible and possible
dQ
> 0, the cycle is impossible, since it violates the second law.
T
[Aliter: Let us consider two heat engine cycles, one reversible and the other irreversible, operating between the same two constant temperature reservoirs T1 and
T2. For the supply of the same quantity of heat Q1 to both the engines, and since the
efficiency of the irreversible engine hI is less than that of the reversible engine hR
hI < hR
WI
W
< R
Q1
Q1
\
WI < WR
or,
Q1 – Q2 < Q1 – Q¢2
\
Q¢2 > Q2
where Q2 is the heat rejection from the reversible engine and Q¢2 is that from the
irrever-sible engine. For the reversible engine,
dQ
Q Q
ÑÚ T = T11 - T22 = 0
For the irreversible engine,
dQ
Q Q
ÑÚ T = T11 - T22 < 0
Similarly, it can be proved that Ñ
Ú
dQ
< 0 for any cyclic device operating beT
tween more than two reservoirs.
dQ
£0
T
The equality sign holds good for any reversible cycle and the inequality is true
for an irreversible cycle. The integration has to be performed over the entire cycle.
So, in general Ñ
Ú
dQ
> 0, that cycle is impossible. The Clausius inequality is also
T
valid for refrigerators and heat pumps.]
If for a cycle, Ñ
Ú
7.7
ENTROPY CHANGE IN AN IRREVERSIBLE PROCESS
For any process undergone by a system, we have from Eq. (7.8)
dQ
T
£ ds
165
Entropy
dQ
ds ≥
or
(7.10)
T
This is further clarified if we consider the cycle as shown in Fig. 7.11, where A
and B are reversible processes and C is an irreversible process. For the reversible
cycle consisting of A and B
Re
2
v.
T
A
B
C
v.
Re
Irre
1
v.
S
Fig. 7.11 Entropy Change in an Irreversible Process
z
z
z
dQ
R
or
T
z
=
2 dQ
+
T
1
1
T
T
2
A
=0
B
-
2 dQ
z
1 dQ
z
=–
A
1 dQ
(7.11)
T
2
B
For the irreversible cycle consisting of A and C, by the inequality of Clausius,
dQ
T
2 dQ
z z
z z
z z
z z
=
+
T
1
A
1 dQ
2
T
1
dQ
2
T
1
dQ
2
T
<0
(7.12)
C
From Eqs. (7.11) and (7.12),
-
1
dQ
2
T
+
B
\
1
<0
C
dQ
T
2
>
B
(7.13)
C
Since the path B is reversible,
1
2
dQ
T
1
=
2
B
dS
(7.14)
B
Since entropy is a property, entropy changes for the paths B and C would be the
same. Therefore,
z z
1
2
B
dS =
1
2
C
dS
(7.15)
166
Engineering Thermodynamics
From Eqs. (7.13) to (7.15),
z z
1
2
1
dQ
2
T
dS >
C
C
Thus, for any irreversible process,
dS >
dQ
T
whereas for a reversible process,
dS =
dQ
rev
T
Therefore, for the general case, we can write,
dS ≥
dQ
T
S2 – S1 ≥
or
z
2 dQ
(7.16)
T
The equality sign holds good for a reversible process and the inequality sign for an
irreversible process.
1
7.8 ENTROPY PRINCIPLE
For any infinitesimal process undergone by a system, we have from Eq. (7.10) for
the total mass
dS ≥
dQ
T
For an isolated system which does not undergo any energy interaction with the
surroundings d-Q = 0.
Therefore, for an isolated system
dSiso ≥ 0
(7.17)
For a reversible process,
dSiso = 0
or
S = constant
For an irreversible process
dSiso > 0
It is thus proved that the entropy of an isolated system can never decrease. It
always increases and remains constant only when the process is reversible. This is
known as the principle of increase of entropy, or simply the entropy principle. It is
the quantitative general statement of second law from the macroscopic viewpoint.
167
Entropy
An isolated system can always be formed by including any system and its surroundings within a
single boundary (Fig. 7.12). Sometimes the original
system which is then only a part of the isolated system is called a ‘subsystem’.
The system and the surroundings together (the
universe or the isolated system) include everything
which is affected by the process. For all possible
processes that a system in the given surroundings
can undergo
System
W
Q
Surroundings
Isolated (composite) system
Fig. 7.12 Isolated System
dSuniv ≥ 0
or
dSsys + dSsurr ≥ 0
(7.18)
Entropy may decrease locally at some region within the isolated system, but it must
be compensated by a greater increase of entropy somewhere within the system so
that the net effect of an irreversible process is an entropy increase of the whole
system. The entropy increase of an isolated system is a measure of the extent of
irreversibility of the process undergone by the system.
Rudolf Clausius summarized the first and second laws of thermodynamics in the
following words:
(a) Die Energie der Welt ist konstant.
(b) Die Entropie der Welt strebt einem Maximum zu.
[(a) The energy of the world (universe) is constant.
(b) The entropy of the world tends towards a maximum.]
The entropy of an isolated system always increases and becomes a maximum at
the state of equilibrium. If the entropy of an isolated system varies with some parameter x, then there is a certain value of xe which maximizes the entropy
FG when d S = 0IJ and represents the equilibrium state (Fig. 7.13). The system is then
H dx K
said to exist at the peak of the entropy hill, and dS = 0. When the system is at
equilibrium, any conceivable change in entropy would be zero.
Siso
Smax
Equilibrium
state
Xe
X
Fig. 7.13
Equilibrium State of an Isolated System
168
7.9
Engineering Thermodynamics
APPLICATIONS OF ENTROPY PRINCIPLE
The principle of increase of entropy is one of the most important laws of physical science. It is the quantitative statement of the second law of thermodynamics. Every irreversible process is accompanied by entropy increase of the universe, and this entropy
increase quantifies the extent of irreversibility of the process. The higher the entropy
increase of the universe, the higher will be the irreversibility of the process. A few
applications of the entropy principle are illustrated in the following.
7.9.1
Transfer of Heat through a Finite Temperature Difference
Let Q be the rate of heat transfer from reservoir A at T1 to reservoir B at
T2, T1 > T2 (Fig. 7.14).
System boundary
Q
T1
T2
Reservoir A
Reservoir B
Fig. 7.14
Heat Transfer through a Finite Temperature Difference
For reservoir A, DSA = – Q/ T1 . It is negative because heat Q flows out of the
reservoir. For reservoir B, DSB = + Q/T2. It is positive because heat flows into the
reservoir. The rod connecting the reservoirs suffers no entropy change because,
once in the steady state, its coordinates do not change.
Therefore, for the isolated system comprising the reservoirs and the rod, and
since entropy is an additive property
S = SA + SB
D Suniv = DSA + D SB
or
DSuniv = -
Q
T1
+
Q
T2
=Q◊
T1 - T2
T2T2
Since T1 > T2, D Suniv is positive, and the process is irreversible and possible. If T1 =
T2, D Suniv is zero, and the process is reversible. If T1 < T2, DSuniv is negative and the
process is impossible.
7.9.2
Mixing of Two Fluids
Subsystem 1 having a fluid of mass m1, specific heat c1, and temperature t 1, and
subsystem 2 consisting of a fluid of mass m2, specific heat c2, and temperature t2 ,
comprise a composite system in an adiabatic enclosure (Fig. 7.15). When the partition is removed, the two fluids mix together, and at equilibrium let tf be the final
temperature, and t2 < tf < t 1. Since energy interaction is exclusively confined to the
two fluids, the system being isolated
m1c1 (t1 – tf ) = m2c2 (tf – t 2)
169
Entropy
Partition
Adiabatic
enclosure
m1
m2
c2
c1
t1
t1 > t2
t2
Subsystem 2
Subsystem 1
Mixing of Two Fluids
Fig. 7.15
\
tf =
m1 c 1 t 1 + m 2 c 2 t 2
m1 c 1 + m 2 c 2
Entropy change for the fluid in subsystem 1
D S1 =
Tf
d-Q rev
T1
T
z
= m1c1 ln
=
z
Tf
m1c 1d T
Ti
T
= m1c 1 ln
Tf
T1
t f + 273
t 1 + 273
This will be negative, since T1 > Tf
Entropy change for the fluid in subsystem 2
Tf m c dT
T
t + 273
2 2
D S2 =
= m2c2 ln f = m2c2 ln f
T2
T2
t 2 + 273
T
z
\
This will be positive, since T2 < Tf
D Suniv = D S1 + D S2
Tf
Tf
= m1c1 ln
+ m2c 2 ln
T1
T2
D Suniv will be positive definite, and the mixing process is irreversible.
Although the mixing process is irreversible, to evaluate the entropy change for
the subsystems, the irreversible path was replaced by a reversible path on which the
integration was performed.
If m1 = m2 = m and c1 = c2 = c.
D Suniv = mc ln
Tf2
T1 ◊ T2
m1c1T1 + m 2 c 2T2
and
Tf =
\
D Suniv = 2mc ln
m1c1 + m 2c 2
=
(T1 + T2 )/ 2
T1 ◊ T2
T1 + T2
2
170
Engineering Thermodynamics
This is always positive, since the arithmetic mean of any two numbers is always
greater than their geometric mean. This can also be proved geometrically. Let a
semi-circle be drawn with (T1 + T2) as diameter (Fig. 7.16).
E
A
O
T1
D
B
C
T2
Fig. 7.16 Geometrical Proof to Show That g.m. < a.m.
Here, AB = T1, BC = T2 and OE = (T1 + T2)/2. It is known that (DB)2 =
AB ◊ BC = T1T2.
\
DB = T1T2
Now,
OE > DB
T1 + T2
2
> T1T2
7.9.3 Maximum Work Obtainable from Two Finite Bodies at
Temperatures T1 and T2
Let us consider two identical finite bodies of constant heat capacity at temperatures
T1 and T2 respectively, T1 being higher than T2. If the two bodies are merely brought
together into thermal contact, delivering no work, the final temperature Tf reached
would be the maximum
Tf =
T1 + T2
2
If a heat engine is operated between the
two bodies acting as thermal energy reservoirs (Fig. 7.17), part of the heat withdrawn
from body 1 is converted to work W by the
heat engine, and the remainder is rejected to
body 2. The lowest attainable final temperature Tf corresponds to the delivery of the
largest possible amount of work, and is associated with a reversible process.
As work is delivered by the heat engine,
the temperature of body 1 will be decreasing
and that of body 2 will be increasing. When
Body 1
T1 Tf
Q1
W = Q1 - Q 2
H.E.
Q2
Body 2
T2 Tf
Fig. 7.17 Maximum Work Obtainable
from Two Finite Bodies
171
Entropy
both the bodies attain the final temperature Tf , the heat engine will stop operating.
Let the bodies remain at constant pressure and undergo no change of phase.
Total heat withdrawn from body 1
Q1 = Cp (T1 – Tf )
where Cp is the heat capacity of the two bodies at constant pressure
Total heat rejected to body 2
Q2 = Cp (Tf – T2)
\ Amount of total work delivered by the heat engine
W = Q1 – Q2
= Cp (T1 + T2 – 2Tf )
(7.19)
For given values of Cp, T1 and T2, the magnitude of work W depends on Tf . Work
obtainable will be maximum when Tf is minimum.
Now, for body 1, entropy change D S1 is given by
D S1 =
z
Tf
T1
Cp
dT
T
= Cp ln
Tf
T1
For body 2, entropy change DS2 would be
D S2 =
z
Tf
T2
Cp
dT
T
Tf
= Cp ln T
2
Since the working fluid operating in the heat engine cycle does not undergo any
entropy change, DS of the working fluid in heat engine =
z
dS = 0. Applying the
entropy principle
DSuniv ≥ 0
\
Cp ln
Cp ln
Tf
T1
+ Cp ln
T f2
T1 T2
Tf
T2
≥0
≥0
(7.20)
From Eq. (7.20), for Tf to be a minimum
Cp ln
Tf2
T1 ◊ T2
T f2
or
ln
\
Tf = T1 ◊ T2
T1 T2
=0
= 0 = ln 1
(7.21)
172
Engineering Thermodynamics
For W to be a maximum, Tf will be T1 T2 . From Equation (7.19)
Wmax = Cp (T1 + T2 – 2
T1 T2 ) = Cp ( T1 - T2 ) 2
The final temperatures of the two bodies, initially at T1 and T2, can range from
(T1 + T2 )/2 with no delivery of work to T1 T2 with maximum delivery of work.
7.9.4
Maximum Work Obtainable from a Finite Body and a TER
Let one of the bodies considered in the previous section be a thermal energy reservoir. The finite body has a thermal capacity Cp and is at temperature T and the TER
is at temperature T0, such that T > T0. Let a heat
engine operate between the two (Fig. 7.18). As
Body T
heat is withdrawn from the body, its temperature
decreases. The temperature of the TER would,
Q
however, remain unchanged at T0. The engine
would stop working, when the temperature of the
HE
W
body reaches T0. During that period, the amount
of work delivered is W, and the heat rejected to
Q-W
the TER is (Q – W ). Then
T0
dT
T0
TER
D SBody =
Cp
= Cp ln
T
T0
T
T
DSHE =
z
z
DSTER =
\
Fig. 7.18 Maximum Work
Obtainable When One
of the Bodies is a TER
dS = 0
Q-W
T0
DSuniv = Cp ln
T0
T
+
Q-W
T0
By the entropy principle,
or
or
or
\
or,
DSuniv ≥ 0
T
Q-W
Cp ln 0 +
≥0
T
T0
T0 W - Q
Cp ln
≥
T
T0
T0
W -Q
£ Cp ln
T
T0
T0
W £ Q + T0 Cp ln
T
Wmax = Q + T0Cp ln
Wmax
T0
T
L
TO
= C M(T - T ) - T ln P
T PQ
MN
p
0
0
0
(7.22)
173
Entropy
7.9.5
Processes Exhibiting External Mechanical Irreversibility
(i) Isothermal Dissipation of Work
Let us consider the isothermal dissipation
of work through a system into the internal energy of a reservoir, as in the flow of an
electric current I through a resistor in contact with a reservoir (Fig. 7.19). At steady
W
state, the internal energy of the resistor and
C.V.
hence its temperature is constant. So, by
I
I
first law
W=Q
The flow of current represents work transfer. At steady state the work is dissipated
isothermally into heat transfer to the surroundings. Since the surroundings absorb
Q units of heat at temperature T,
DSsurr =
At steady state,
\
Q
T
=
Q
Surr.
at T
Fig. 7.19 External Mechanical
Irreversibility
W
T
D Ssys = 0
D Suniv= DSsys + D Ssurr =
W
(7.23)
T
The irreversible process is thus accompanied by an entropy increase of the
universe.
(ii) Adiabatic Dissipation of Work Let W be the stirring work supplied to a
viscous thermally insulated liquid, which is dissipated adibatically into internal
energy increase of the liquid, the temperature of which increases from Ti to Tf
(Fig. 7.20). Since there is no flow of heat to or from the surroundings,
DSsurr = 0
f
W
T
p=c
i
Insulation
S
(a)
(b)
Fig. 7.20
Adiabatic Dissipation of Work
174
Engineering Thermodynamics
To calculate the entropy change of the system, the original irreversible path (dotted
line) must be replaced by a reversible one between the same end states, i and f. Let us
replace the irreversible performance of work by a reversible isobaric flow of heat from
a series of reservoirs ranging from Ti to Tf to cause the same change in the state of the
system. The entropy change of the system will be
D Ssys =
z
R
i
f
dQ
T
=
z
R
f C dT
p
i
T
= Cp ln
Tf
Ti
where Cp is the heat capacity of the liquid.
DSuniv = DSsys + DSsurr = Cp ln
Tf
Ti
(7.24)
which is positive.
7.10 ENTROPY TRANSFER WITH HEAT FLOW
dQ
rev
, when heat is added to a system, d-Q is positive, and the entropy
T
of the system increases. When heat is removed from the system, d-Q is negative,
and the entropy of the system decreases.
Heat transferred to the system of
fixed mass increases the internal energy
of the system, as a result of which the
molecules (of a gas) move with higher
T0
kinetic energy and collide more
System
frequently, and so the disorder in the sysSurr
Q
T0
tem increases. Heat is thus regarded as
disorganised or disordered energy transfer which increases molecular chaos (see
Sec. 7.16). If heat Q flows reversibly
Fig. 7.21 Entropy Transfer Along with
from the system to the surroundings at
Heat Flow
T0 (Fig. 7.21, the entropy increase of the
surroundings is
Q
DSsurr =
T0
Since dS =
The entropy of the system is reduced by
Q
DSsys = –
T0
The temperature of the boundary where heat transfer occurs is the constant temperature T0. It may be said that the system has lost entropy to the surroundings.
Alternatively, one may state that the surroundings have gained entropy from the
system. Therefore, there is entropy transfer from the system to the surroundings
along with heat flow. In other words, since the heat inflow increases the molecular
disorder, there is flow of disorder along with heat.
175
Entropy
On the other hand, there is no entropy transfer associated with work. In
Fig. 7.22, the system delivers work to a flywheel, where energy is stored in a fully
recoverable form. The flywheel molecules are simply put into rotation around the
axis in a perfectly organised manner, and there is no dissipation and hence no entropy increase of the flywheel. The same can be said about work transfer in the
compression of a spring or in the raising of a weight by a certain height. There is
thus no entropy transfer along with work. If work is dissipated adiabatically into
internal energy increase of the system (Sub-section 7.9.5), there is an entropy increase in the system, but there is as such no entropy transfer to it.
Flywheel
W
System
Fig. 7.22
No Entropy Transfer Along with Work Transfer
7.11 ENTROPY GENERATION IN A CLOSED SYSTEM
The entropy of any closed system can increase in two ways:
(a) by heat interaction in which there is entropy transfer
(b) by internal irreversibilities or dissipative effects in which work (or K.E.) is
dissipated into internal energy increase.
If d-Q is the infinitesimal amount of heat transferred to the system through its
boundary at temperature T, the same as that of the surroundings, the entropy
increase dS of the system can be expressed as
dS = deS + diS
=
dQ
+ diS
(7.25)
T
where deS is the entropy increase due to external heat interaction and diS is the
entropy increase due to internal irreversibility. From Eq. (7.25),
dS ≥
\
dQ
T
diS ≥ 0
(7.26)
The entropy increase due to internal irreversibility is also called entropy
production or entropy generation, Sgen.
It may so happen that in a process (e.g. the expansion of a hot fluid in a turbine)
dQ
the entropy decrease of the system due to heat loss to the surroundings is
T
F
GH
z IJK
176
Engineering Thermodynamics
equal to the entropy increase of the system due to internal irreversibilities such as
z
e d Sj , in which case the entropy of the system before and after the
process will remain the same e dS = 0j . Therefore, an isentropic process need not
friction, etc.
i
z
be adiabatic or reversible. But if the isentropic process is reversible, it must be
adiabatic. Also, if the isentropic process is adiabatic, it cannot but be reversible.
An adiabatic process need not be isentropic, since entropy can also increase due to
friction etc. But if the process is adiabatic and reversible, it must be isentropic.
For an infinitesimal reversible process by a closed system,
d-QR = dUR + pdV
If the process is irreversible,
d-QI = d-UI + d-W
Since U is a property,
dUR = dUI
\
d-QR – pdV = d-QI – d- W
F dQ I = FG d QIJ + pdV - d W
GH T JK H T K
T
-
-
or
(7.27)
I
R
The difference (pdV – dW ) indicates the work that is lost due to irreversibility,
and is called the lost work d- (L W ), which approaches zero as the process approaches reversibility as a limit. Equation (7.27) can be expressed in the form
dS = deS + diS
Thus the entropy of a closed system increases due to heat addition (deS) and
internal dissipation (diS).
In any process executed by a system, energy is always conserved, but entropy is
produced internally. For any process between equilibrium states 1 and 2 (Fig. 7.23),
the first law can be written as follows:
z z
2
dQ
1
or
2
1
d- W = E 2 - E1
Energy
transfer
Energy
change
Q1–2 = E2 – E1 + W1–2
By the second law,
S2 – S1 ≥
z
2
d- Q
T
It is only the transfer of energy as heat which is accompanied by entropy transfer,
both of which occur at the boundary where the temperature is T. Work interaction is
1
177
Entropy
Boundary
Surroundings
System
2
1
T
(Boundary
Temperature)
dQ
(Heat Transfer)
dW
(Work Transfer)
dQ (Entropy Transfer)
T
Fig. 7.23 Schematic of a Closed System Interacting with Its Surroundings
not accompanied by any entropy transfer. The entropy change of the system
2 dQ
(S2 – S1) exceeds the entropy transfer
. The difference is produced inter1
T
nally due to irreversibility. The amount of entropy generation Sgen is given by
2 dQ
S2 – S1 –
= Sgen
(7.28)
1
T
z
z
Entropy
change
\
Entropy
transfer
Entropy
production
Sgen ≥ 0
The second law states that, in general, any thermodynamic process is accompanied by entropy generation.
Process 1 – 2, which does not generate any entropy (Sgen = 0), is a reversible
process (Fig. 7.24). Paths for which Sgen > 0 are considered irreversible. Like heat
transfer and work transfer during the process 1 –2, the entropy generation also
depends on the path the system follows. Sgen is, therefore, not a thermodynamic
property and d-Sgen is an inexact differential, although (S2 – S1) depends only on the
end states. In the differential form,
Eq. (7.28) can be written as
d-Sgen = dS –
dQ
(7.29)
T
The amount of entropy generation
quantifies the intrinsic irreversibility of
the process. If the path A causes more
entropy generation than path B
(Fig. 7.24), i.e.
(Sgen)A > (Sgen )B
the path A is more irreversible than
path B and involves more ‘lost work’.
Fig. 7.24 Entropy Generation Depends on
the Path
178
Engineering Thermodynamics
If heat transfer occurs at several locations on the boundary of a system, the en
tropy transfer term can be expressed as a sum, so Eq. (7.28) takes the form
S2 – S1 = S
j
Qj
Tj
+ Sgen
(7.30)
where Q j /T j is the amount of entropy transferred through the portion of the boundary at temperature T j.
On a time rate basis, the entropy balance can be written as
dS
dt
∑
=S
j
Qj
Tj
∑
+ Sgen
(7.31)
∑
where dS/dt is the rate of change of entropy of the system, Qj /Tj is the rate of
entropy transfer through the portion of the boundary whose instantaneous temperature is Tj, and Sgen is the rate of entropy generation due to irreversibilities within the
system.
∑
7.12 ENTROPY GENERATION IN AN OPEN SYSTEM
In an open system, there is transfer of three quantities: mass, energy and entropy.
The control surface can have one or more openings for mass transfer (Fig. 7.25). It
is rigid, and there is shaft work transfer across it.
Surroundings
me
mi
Control
volume
M
E
C.S.
S
Surface
Temperature, T
(Entropy Transfer
Rate)
Fig. 7.25
Wsh
Q
Q
T
(Shaft Work
Transfer Rate)
(Heat Transfer
Rate)
Schematic of an Open System and Its Interaction with Surroundings
The continuity equation gives
S m1 - S m e =
∑
i
∑
e
net mass
transfer rate
∂M
∂t
rate of mass
accumulation
in the CV
(7.32)
179
Entropy
The energy equation gives
FG
H
S m1 h +
∑
i
IJ
K
FG
H
IJ
K
∂E
V2
V2
+ gZ - S m e h +
+ gZ + Q - Wsh =
e
∂t
2
2
i
e
∑
∑
net rate of energy
transfer
∑
(7.33)
rate of energy accumulation
in the CV
The second law inequality or the entropy principle gives
∑
Q ∂S
S m i Si - S m e Se + £
i
e
T ∂t
net rate of entropy
rate of increase of
transfer
entropy of the CV
∑
∑
(7.34)
∑
Here Q represents the rate of heat transfer at the location of the boundary where the
∑
instantaneous temperature is T. The ratio Q / T accounts for the entropy transfer
along with heat. The terms m i Si and m e Se account, respectively, for rates of
entropy transfer into and out of the CV accompanying mass flow. The rate of entropy
increase of the control volume exceeds, or is equal to, the net rate of entropy transfer
into it. The difference is the entropy generated within the control volume due to
irreversibility. Hence, the rate of entropy generation is given by
∑
∑
∑
∂S
Q
- S m i Si + S m e Se Sgen =
i
e
∂t
T
∑
∑
∑
(7.35)
By the second law,
∑
Sgen ≥ 0
∑
∑
If the process is reversible, Sgen = 0. For an irreversible process, Sgen > 0. The
∑
magnitude of Sgen quantifies the irreversibility of the process. If systems A and B
∑
∑
operate so that ( Sgen )A > ( Sgen )B it can be said that the system A operates more
∑
irreversibly than system B. The unit of Sgen is W/K.
As steady state, the continuity equation gives
S mi = S m e
∑
∑
i
(7.36)
e
the energy equation becomes
∑
FG
H
0 = Q - Wsh + S m i h +
∑
i
IJ
K
FG
H
IJ
K
V2
V2
+ gZ - S m e h +
+ gZ
e
2
2
i
e
∑
(7.37)
and the entropy equation reduces to
∑
Q
0 = + S m i Si - S m e se + Sgen
e
T i
∑
∑
∑
(7.38)
180
Engineering Thermodynamics
These equations often must be solved simultaneously, together with appropriate
property relations.
Mass and energy are conserved quantities, but entropy is not generally
conserved. The rate at which entropy is transferred out must exceed the rate at
which entropy enters the CV, the difference being the rate of entropy generated
within the CV owing to irreversibilities.
For one-inlet and one-exit control volumes, the entropy equation becomes
∑
Q
0 = + m( s1 - s2 ) + Sgen
T
∑
F I
GG JJ
H K
∑
\
s2 – s1 =
∑
1 Q
S gen
+
m
m T
∑
(7.39)
7.13 FIRST AND SECOND LAWS COMBINED
By the second law
_
d Qrev = TdS
and by the first law, for a closed non-flow system,
_
d Q = dU + pdV
TdS = dU + pdV
(7.40)
Again, the enthalpy
H = U + pV
dH = dU + pdV + Vdp
= TdS + Vdp
TdS = dH – Vdp
(7.41)
Equations (7.40) and (7.41) are the thermodynamic equations relating the
properties of the system.
Let us now examine the following equations as obtained from the first and second
laws:
_
_
(a) d Q = dE + d W—This equation holds good for any process, reversible or
irreversible, and for any system.
_
_
(b) d Q = dU + d W—This equation holds good for any process undergone by a
closed stationary system.
_
(c) d Q = dU + pdV—This equation holds good for a closed system when only
pdV-work is present. This is true only for a reversible (quasi-static) process.
_
(d) d Q = TdS—This equation is true only for a reversible process.
(e) TdS = dU + pdV—This equation holds good for any process reversible or
irreversible, undergone by a closed system, since it is a relation among properties which are independent of the path.
(f) TdS = dH – Vdp—This equation also relates only the properties of the closed
system. There is no path function term in the equation. Hence the equation
holds good for any process.
181
Entropy
The use of the term ‘irreversible process’ is doubtful, since no irreversible path
or process can be plotted on thermodynamic coordinates. It is more logical to state
that ‘the change of state is irreversible’ rather than say ‘it is an irreversible process’.
A natural process which is inherently irreversible is indicated by a dotted line
connecting the initial and final states, both of which are in equilibrium. The dotted
line has no other meaning, since it can be drawn in any way. To determine the
entropy change for a real process, a known reversible path is made to connect the
two end states, and integration is performed on this path using either equation (e) or
equation (f), as given above. Therefore, the entropy change of a system between
two identifiable equilibrium states is the same whether the intervening process is
reversible or the change of state is irreversible.
7.14 REVERSIBLE ADIABATIC WORK IN A STEADY FLOW
SYSTEM
In the differential form, the steady flow energy equation per unit mass is given by
Eq. (5.11),
_
_
d Q = dh + VdV + gdZ + d Wx
_
For a reversible process, d Q = Tds
_
\
Tds = dh + VdV + gdZ + d Wx
(7.42)
Using the property relation, Eq. (7.41), per unit mass,
Tds = dh – vdp
in Eq. (7.42), we have
_
– vdp = VdV + gdZ + d Wx
(7.43)
z
(7.44)
On integration
V2
+ gD Z + Wx
1
2
If the changes in K.E. and P.E. are neglected, Eq. (7.44) reduces to
–
2
vdp = D
z
Wx = –
2
(7.45)
vdp
1
_
If d Q = 0, implying ds = 0, the property relation gives
dh = vdp
or
h2 – h1 =
z
2
(7.46)
vdp
1
From Eqs. (7.45) and (7.46),
Wx = h1 – h2 = –
The integral –
z
z
2
vdp
(7.47)
1
2
1
vdp represents an area on the p–v plane (Fig. 7.26). To make
the integration, one must have a relation between p and v such as pvn = constant.
\
182
Engineering Thermodynamics
W1– 2 = h1 – h2 = –
z
2
vdp
1
b
= area 12 ab 1
W1 – 2 =
z
dp
Equation (7.47) holds good for
a steady flow work-producing
machine like an engine or
turbine as well as for a workabsorbing machine like a pump
or a compressor, when the fluid
undergoes reversible adiabatic
expansion or compression.
It may be noted that for a
closed stationary system like a
gas confined in a piston-cylinder
machine (Fig. 7.27a), the
reversible work done would be
1
2
p
Wx =
Ú v dp
1
v
2
a
v
Fig. 7.26
Reversible Steady Flow Work
Interaction
2
pdV = Area 12 cd 1
1
Closed system
C.S.
1
C.V.
2
1
1
b
2
p
2
p
W1–2 = Ú p dv
W1–2 =
1
Ú vdp
1
2
a
2
d
v
(a)
Fig. 7.27
c
v
(b)
Reversible Work Transfer in (a) A Closed System and (b) A
Steady Flow System
Entropy
183
The reversible work done by a steady flow system (Fig. 7.27b) would be
W1–2 = –
z
2
vdp = Area 12 ab 1
1
7.15 ENTROPY AND DIRECTION: THE SECOND LAW—A
DIRECTIONAL LAW OF NATURE
Since the entropy of an isolated system can never decrease, it follows that only those
processes are possible in nature which would give an entropy increase for the system
and the surroundings together (the universe). All spontaneous processes in nature
occur only in one direction from a higher to a lower potential, and these are
accompanied by an entropy increase of the universe. When the potential gradient is
infinitesimal (or zero in the limit), the entropy change of the universe is zero, and the
process is reversible. The second law indicates the direction in which a process takes
place. A process always occurs in such a direction as to cause an increase in the
entropy of the universe. The macroscopic change ceases only when the potential
gradient disappears and the equilibrium is reached when the entropy of the universe
assumes a maximum value. To determine the equilibrium state of an isolated system
it is necessary to express the entropy as a function of certain properties of the system
and then render the function a maximum. At equilibrium, the system (isolated) exists
at the peak of the entropy-hill, and dS = 0 (Fig. 7.13).
7.16
ENTROPY AND DISORDER
Work is a macroscopic concept. Work involves order or the orderly motion of
molecules, as in the expansion or compression of a gas. The kinetic energy and
potential energy of a system represent orderly forms of energy. The kinetic energy
of a gas is due to the coordinated motion of all the molecules with the same average
velocity in the same direction. The potential energy is due to the vantage position
taken by the molecules or displacements of molecules from their normal positions.
Heat or thermal energy is due to the random thermal motion of molecules in a
completely disorderly fashion and the average velocity is zero. Orderly energy can
be readily converted into disorderly energy, e.g. mechanical and electrical energies
into internal energy (and then heat) by friction and Joule effect. Orderly energy can
also be converted into one another. But there are natural limitations on the
conversion of disorderly energy into orderly energy, as delineated by the second
law. When work is dissipated into internal energy, the disorderly motion of
molecules is increased. Two gases, when mixed, represent a higher degree of
disorder than when they are separated. An irreversible process always tends to take
the system (isolated) to a state of greater disorder. It is a tendency on the part of
nature to proceed to a state of greater disorder. An isolated system always tends to
a state of greater entropy. So there is a close link between entropy and disorder. It
may be stated roughly that the entropy of a system is a measure of the degree of
molecular disorder existing in the system. When heat is imparted to a system, the
disorderly motion of molecules increases, and so the entropy of the system
increases. The reverse occurs when heat is removed from the system.
184
Engineering Thermodynamics
Ludwig Boltzmann (1877) introduced statistical concepts to define disorder by
attaching to each state a thermodynamic probability, expressed by the quantity W,
which is greater the more disordered the state is. The increase of entropy implies
that the system proceeds by itself from one state to another with a higher
thermodynamic probability (or disorder number). An irreversible process goes on
until the most probable state (equilibrium state when W is maximum) corresponding
to the maximum value of entropy is reached. Boltzmann assumed a functional
relation between S and W. While entropy is additive, probability is multiplicative.
If the two parts A and B of a system in equilibrium are considered, the entropy is the
sum
S = SA + SB
and the thermodynamic probability is the product
W = WA ◊ WB
Again, S = S(W ), SA = S(WA), and SB = S(WB)
\
S(W) = S(WAWB) = S(WA) + S(WB)
which is a well-known functional equation for the logarithm. Thus the famous
relation is reached
S = K ln W
where K is a constant, known as Boltzmann constant. This is engraved upon
Boltzmann’s tombstone in Vienna.
When W = 1, which represents the greatest order, S = 0. This occurs only at
T = 0 K. This state cannot be reached in a finite number of operations. This is the
Nernst-Simon statement of third law of thermodynamics. In the case of a gas, W
increases due to an increase in volume V or temperature T. In the reversible adiabatic
expansion of a gas the increase in disorder due to an increase in volume is just
compensated by the decrease in disorder due to a decrease in temperature, so that
the disorder number or entropy remains constant.
7.17
ABSOLUTE ENTROPY
It is important to note that one is interested only in the amount by which the entropy
of the system changes in going from an initial to final state, and not in the value of
absolute entropy. In cases where it is necessary, a zero value of entropy of the
system at an arbitrarily chosen standard state is assigned, and the entropy changes
are calculated with reference to this standard state.
7.18
POSTULATORY THERMODYNAMICS
The property ‘entropy’ plays the central role in thermodynamics. In the classical
approach, as followed in this book, entropy is introduced via the concept of the
heat engine. It follows the way in which the subject of thermodynamics developed
historically mainly through the contributions of Sadi Carnot, James Prescott Joule,
William Thomson (Lord Kelvin), Rudolf Clausius, Max Planck and Josiah Willard
Gibbs.
185
Entropy
In the postulatory approach, as developed by HB Callen (see Reference),
entropy is introduced at the beginning. The development of the subject has been
based on four postulates. Postulate I defines the equilibrium state. Postulate II
introduces the property ‘entropy’ which is rendered maximum at the final
equilibrium state. Postulate III refers to the additive nature of entropy which is a
monotonically increasing function of energy. Postulate IV mentions that the entropy
of any system vanishes at the absolute zero of temperature. With the help of these
postulates the conditions of equilibrium under different constraints have been
developed.
Solved Examples
Example 7.1 Water flows through a turbine in which friction causes the water
temperature to rise from 35°C to 37°C. If there is no heat transfer, how much does
the entropy of the water change in passing through the turbine? (Water is
incompressible and the process can be taken to be at constant volume.)
Solution The presence of friction makes the process irreversible and causes an
entropy increase for the system. The flow process is indicated by the dotted line
joining the equilibrium states 1 and 2 (Fig. 7.28). Since entropy is a state property
and the entropy change depends only on the two end states and is independent of
the path the system follows, to find the entropy change, the irreversible path has to
be replaced by a reversible path, as shown in the figure, because no integration can
be made on a path other than a reversible path.
Irreversible
310 K
2
T
Reversible path
connecting the initial
and final equilibrium
states
T
308 K
1
dQrev
ds
s
Fig. 7.28
T 2 = 37 + 273 = 310 K
T1 = 35 + 273 = 308 K
We have
_
d Q rev = TdS
dS =
mcv dT
T
186
Engineering Thermodynamics
S2 – S1 = mcv ln
T2
T1
310
= 0.0243 kJ/K
308
Example 7.2 (a) One kg of water at 273 K is brought into contact with a heat
reservoir at 373 K. When the water has reached 373 K, find the entropy change of
the water, of the heat reservoir, and of the universe.
(b) If water is heated from 273 to 373 K by
first bringing it in contact with a reserReservoir
voir at 323 K and then with a reservoir at
373 K
373 K, what will the entropy change of the
universe be?
(c) Explain how water might be heated from
Q
273 to 373 K with almost no change in
the entropy of the universe.
= 1 ¥ 4.187 ln
System
Solution (a) Water is being heated through a
273 K
finite temperature difference (Fig. 7.29). The
entropy of water would increase and that of the
reservoir would decrease so that the net entropy
Fig. 7.29
change of the water (system) and the reservoir
together would be positive definite. Water is being heated irreversibly, and to find
the entropy change of water, we have to assume a reversible path between the end
states which are at equilibrium.
_
T1 mcdT
dQ
T
(DS)water =
=
= mc ln 2
T1
T
T1
T
z
z
= 1 ¥ 4.187 ln
373
= 1.305 kJ/K
273
The reservoir temperature remains constant irrespective of the amount of heat
withdrawn from it.
Amount of heat absorbed by the system from the reservoir,
Q = 1 ¥ 4.187 ¥ (373 – 273) = 418.7 kJ
\ Entropy change of the reservoir
418.7
Q
== – 1.122 kJ/K
T
373
\ Entropy change of the universe
(DS)res = -
(DS)univ = (DS)system + (DS)res
= 1.305 – 1.122 = 0.183 kJ/K
(b) Water is being heated in two stages, first from 273 K to 323 K by bringing it
in contact with a reservoir at 323 K, and then from 323 K to 373 K by bringing it in contact of a second reservoir at 373 K.
187
Entropy
(DS)water =
z
323 K
mc
dT
273 K
T
+
z
373 K
mc
323 K
dT
T
FG 323 + ln 373IJ = (0.1673 + 0.1441) 4.187
H 273 323K
= 4.187 ln
= 1.305 kJ/K
(DS)res I = –
f
a
f
1 ¥ 4.187 ¥ 373 - 323
= – 0.56 kJ/K
373
(DS)univ = (DS)water + (DS)res I + (DS)res II
(DS)res II =
\
a
1 ¥ 4.187 ¥ 323 - 273
= – 0.647 kJ/K
323
= 1.305 – 0.647 – 0.56 = 0.098 kJ/K
(c) The entropy change of the universe would be less and less, if the water is
heated in more and more stages, by bringing the water in contact successively
with more and more heat reservoirs, each succeeding reservoir being at a
higher temperature than the preceding one.
When water is heated in infinite steps, by bringing it in contact with an infinite
number of reservoirs in succession, so that at any instant the temperature difference
between the water and the reservoir in contact is infinitesimally small, then the net
entropy change of the universe would be zero, and the water would be reversibly
heated.
Example 7.3 One kg of ice at – 5°C is
exposed to the atmosphere which is at 20°C. The
ice melts and comes into thermal equilibrium
with the atmosphere. (a) Determine the entropy
increase of the universe. (b) What is the
minimum amount of work necessary to convert
the water back into ice at – 5°C? cp of ice is
2.093 kJ/kg K and the latent heat of fusion of ice
is 333.3 kJ/kg.
Atmosphere
20ºC
Q
Ice
– 5ºC
Solution Heat absorbed by ice Q from the
atmosphere (Fig. 7.30).
Fig. 7.30
= Heat absorbed in solid phase + Latent heat
+ Heat absorbed in liquid phase
= 1 ¥ 2.093 ¥ [0 – (– 5)] + 1 ¥ 333.3 + 1 ¥ 4.187 ¥ (20 – 0) = 427.5 kJ
Entropy change of the atmosphere
427.5
Q
(DS)atm = –
=–
= – 1.46 kJ/K
293
T
Entropy change of the system (ice) as it gets heated from – 5°C to 0°C
293
273
dT
(DSI)system =
= 1 ¥ 2.093 ln
= 2.093 ¥ 0.0186
mc p
268
268
T
z
= 0.0389 kJ/K
188
Engineering Thermodynamics
Entropy change of the system as ice melts at 0°C to become water at 0°C
333.3
= 1.22 kJ/K
273
Entropy change of water as it gets heated from 0°C to 20°C
(DSII)system =
z
293
293
dT
= 1 ¥ 4.187 ln
= 0.296 kJ/K
273
T
Total entropy change of ice as it melts into water
(DSIII)system =
273
mc p
(DS)total = DSI + DSII + DSIII
= 0.0389 + 1.22 + 0.296 = 1.5549 kJ/K
The entropy–temperature diagram for the system as ice at – 5°C converts to
water at 20°C is shown in Fig. 7.31.
\ Entropy increase of the universe
(DS)univ = (DS)system + (DS)atm
= 1.5549 – 1.46 = 0.0949 kJ/K
293 K
4
T
(b) To convert 1 kg of water at
20°C to ice at – 5°C, 427.5 kJ
of heat have to be removed
from it, and the system has to
be brought from state 4 to state
1 (Fig. 7.31). A refrigerator
cycle, as shown in Fig. 7.32, is
assumed to accomplished this.
The entropy change of the system
would be the same, i.e. S4 – S1, with
the only difference that its sign will
be negative, because heat is removed
from the system (Fig. 7.31).
(DS)system = S1 – S4
(negative)
273 K
2
268 K
1 10.465 kJ
DS I
1 atm
3
333.3 kJ
83.7 kJ
DS II
DS III
s
Fig. 7.31
The entropy change of the working fluid in the
refrigerator would be zero, since it is operating
in a cycle, i.e
(DS)ref = 0
The entropy change of the atmosphere (positive)
Q+W
T
\ Entropy change of the universe
(DS)univ = (DS)system + (DS)ref + (DS)atm
(DS)atm =
= (S1 – S4) +
Q+W
T
Fig. 7.32
189
Entropy
By the principle of increase of entropy
(DS)univ or
≥0
isolated system
LMbS - S g + Q + W OP ≥ 0
T Q
N
\
1
4
Q+W
≥ (S4 – S1)
T
W ≥ T(S4 – S1) – Q
W(min) = T(S4 – S1) – Q
Q = 427.5 kJ
T = 293 K
\
\
Here
S4 – S1 = 1.5549 kJ/K
\
W(min) = 293 ¥ 1.5549 – 427.5 = 28.5 kJ
Example 7.4 A body of constant heat capacity cp and initial temperature T 1 is
placed in contact with a heat reservoir at temperature T2 and comes to thermal
equilibrium with it. If T 2 > T 1, calculate the entropy change of the universe and
show that this is always positive.
T2
Solution
T
dT
= mcp ln 2
T1
T
T
DSbody = mcp Ú
1
DSreservoir =
-mc p (T2 - T1)
T2
È T
T ˆ˘
Ê
DSuniverse = mcp Íln 2 - Á1 - 1 ˜ ˙
Î T1 Ë T2 ¯ ˚
Putting
Since
1–
1
T1
T
= x, 2 =
1
x
T2
T1
T2
> 1, x is positive.
T1
DS universe
= – ln (1 – x) – x`
mc p
=
\
DSuniverse > 0
Proved.
x 2 x3 x 4
+
+
+ ...
2
3
4
190
Engineering Thermodynamics
Example 7.5 A STP, 8.4 litres of oxygen and 14 litres of hydrogen mix with each
other completely in an insulated chamber. Calculate the entropy change for the
process assuming both gases behave as ideal gases.
Solution
DS = – R (n1 ln x1 + n2 ln x2 )
where
n1 =
8.4
14
= 0.375, n2 =
= 0.625
22.4
22.4
The mole fraction of oxygen, x1 =
n1
= 0.375
n1 + n2
The mole fraction of hydrogen, x2 =
n2
= 0.625
n1 + n2
\
DS = – R (0.375 ln 0.375 + 0.625 ln 0.625)
= 0.66 R = 5.49 J/K Ans.
Example 7.6 A reversible power cycle R and an irreversible power cycle I
operate between the same two reservoirs. Each receives QH from the hot reservoir.
The reversible cycle develops work WR while the irreversible cycle develops work
WI .
(a) Evaluate the rate of entropy generation s for cycle I in terms of WI, WR and
the temperature T C of the cold reservoir. (b) Demonstrate that s must be positive.
Solution
(a) For cycle I,
Hot reservoir TH
d-Q
Q¢ ˘
ÈQ
=–Í H - C ˙
T
TC ˚
Î TH
Energy balances for cycles R and I give
Q H = WR + Q C
QH = WI + Q¢C
\
Q¢C = QC + WR – WI
Therefore,
s = –Ñ
Ú
QH
WR
R
QH
WI
I
QC
Q¢C
Cold reservoir TC
Fig. 7.33
ÈÊ Q
Q ˆ Ê W - WI ˆ ˘
s = – ÍÁ H - C ˜ - Á R
˙
TC ¯ Ë TC ˜¯ ˚
ÎË TH
Since R is reversible,
Q
QH
= C
TH
TC
\
s=
WR - WI
TC
Ans.
Example 7.7 Two identical bodies of constant heat capacity are at the same
initial temperature T i. A refrigerator operates between these two bodies until one
body is cooled to temperature T 2. If the bodies remain at constant pressure and
191
Entropy
undergo no change of phase, show that the
minimum amount of work needed to do this is
FG T + T - 2T IJ
HT
K
2
W(min) = Cp
Ti Æ T 2¢
Body B
Q+W
i
i
2
2
Solution Both the finite bodies A and B are
initially at the same temperature T i. Body A is to
be cooled to temperature T 2 by operating the
refrigerator cycle, as shown in Fig. 7.34. Let T 2¢
be the final temperature of body B.
Heat removed from body A to cool it from T i
to T 2
W
R
Q
Body A
Ti Æ T2
Fig. 7.34
Q = Cp(T i – T 2)
where Cp is the constant pressure heat capacity of the identical bodies A and B.
Heat discharged to body B
= Q + W = Cp (T 2¢ – T i)
Work input, W
= Cp(T 2¢ – T i) – Cp (T i – T2)
= Cp (T2 ¢ + T 2 – 2T i)
Now, the entropy change of body A
DSA =
z
T2
Ti
(1)
Cp
T
dT
= Cp ln 2 (negative)
T
Ti
Cp
dT
T¢
= Cp ln 2 (positive)
Ti
T
The entropy change of body B
DSB =
z
T2 ¢
Ti
Entropy change (cycle) of refrigerant = 0
\ Entropy change of the universe
(DS)univ = DSA + DSB
= Cp ln
T2
T¢
+ Cp ln 2
Ti
Ti
By the entropy principle
(DS)univ ≥ 0
FG C ln T + C ln T ¢ IJ ≥ 0
T K
H T
2
p
2
p
i
Cp ln
i
T2 T2 ¢
≥ 0
Ti 2
(2)
192
Engineering Thermodynamics
In Eq. (1) with Cp, T2, and Ti being given, W will be a minimum when T2¢ is a
minimum. From Eq. (2), the minimum value of T 2¢ would correspond to
Cp ln
\
T2 T2 ¢
= 0 = ln 1
Ti2
T2 ¢ =
Ti2
T2
From Eq. (1)
FG T + T - 2T IJ
HT
K
2
W(min) = Cp
i
2
i
2
Example 7.8 Three identical finite bodies of constant heat capacity are at
temperatures 300 K, 300 K, and 100 K. If no work or heat is supplied from outside,
what is the highest temperature to
which any one of the bodies can be
C
A
raised by the operation of heat
Tf
Tf
300 K
300 K
engines or refrigerators?
W
R
HE
Q2
B
100 K
(DS)C = C ln
Fig. 7.35
Tf
100
Tf ¢
300
(DS)H.E. = 0
(DS)ref = 0
where C is the heat capacity of each of the three bodies.
(DS)univ ≥ 0
Since
FG C ln T + C ln T + C ln T ¢ IJ ≥ 0
H 300 100
300 K
f
f
C ln
Q3
Tf
Tf
(DS)A = C ln
300
(DS)B = C ln
Q4
Q1
Solution Let the three identical
bodies A, B, and C having the same
heat capacity C be respectively at
300 K, 100 K and 300 K initially,
and let us operate a heat engine and
a refrigerator, as shown in Fig. 7.35.
Let T f be the final temperature of
bodies A and B, and T f¢ be the final
temperature of body C. Now
Tf2 Tf¢
9,000,000
f
≥0
193
Entropy
Tf2 Tf¢
For minimum value of T f C ln
9 ¥ 106
¢
\
Now
= 0 = ln 1
T f2 T f ¢ = 9,000,000
(3)
Q1 = C (300 – Tf)
Q2 = C (T f – 100)
Q4 = C (T f¢ – 300)
Again
Q1 = Heat removed from body A
= Heat discharged to bodies B and C
= Q2 + Q 4
\
C (300 – Tf) = C (T f – 100) + C (Tf¢ – 300)
\
Tf¢ = 700 – 2T f
(4)
Tf¢ will be the highest value when T f is the minimum.
From Eqs. (3) and (4)
Tf2 (700 – 2T f ) = 9,000,000
\
2T f3 – 700 T f2 + 9,000,000 = 0
or
Tf = 150 K
From Eq. (4)
Tf¢ = (700 – 2 ¥ 150) K = 400 K
Ans.
A system has a heat capacity at constant volume
Example 7.9
CV = AT2
where A = 0.042 J/K3.
The system is originally at 200 K, and a thermal reservoir at 100 K is available.
What is the maximum amount of work that can be recovered as the system is cooled
down to the temperature of the reservoir?
Heat removed from the system (Fig. 7.36)
Solution
Q1 =
z
T2
T1
CV dT =
LM T OP
MN 3 PQ
z
T2 = 100 K
T1 = 200 K
System
T1 = 200 K
0.042 T 2 dT
100 K
Q1
3 100 K
= 0.042
=
HE
200 K
0.042
J/K3 (1003 – 2003)K3 = – 98 ¥ 103 J
3
(D S)system =
z
100 K
200 K
W
CV
dT
=
T
z
100 K
200 K
0.042 T 2
dT
T
Q1 – W
Reservoir
100 K
Fig. 7.36
194
Engineering Thermodynamics
=
0.042
J/K3 [1002 – 2002]K2 = – 630 J/K
2
(DS)res =
98 ¥ 103 - W
Q1 - W
=
J/K
100
Tres
(DS)working fluid in H.E. = 0
\
(DS)univ = (DS)system + (DS)res
= – 630 +
(DS)univ ≥ 0
Since
\
98 ¥ 103 - W
100
– 630 +
98 ¥ 103 - W
≥ 0
100
980 –
W
– 630 ≥ 0
100
W
£ 350
100
W(max) = 35,000 J = 35 kJ
Example 7.10 A fluid undergoes a reversible adiabatic compression from
0.5 MPa, 0.2 m3 to 0.05 m3 according to the law, pv1.3 = constant. Determine the
change in enthalpy, internal energy and entropy, and the heat transfer and work
transfer during the process.
Solution
2
TdS = dH – Vdp
PV1-3 = Const
p
For the reversible adiabatic process (Fig. 7.37)
dH = Vdp
p1 = 0.5 MPa,
V1 = 0.2 m3
V2 = 0.05 m 3,
p1 V n1 = p2 V n2
FV I
p = p G J
HV K
0.20 I
= 0.5 ¥ F
H 0.05K MPa
F 0.20I MPa
= 0.5 ¥
H 0.05K
1
n
\
2
1
2
1.3
1.3
= 0.5 ¥ 6.061 MPa = 3.0305 MPa
p1V n1 = pV n
v
1
Fig. 7.37
195
Entropy
F p V IJ
V= G
H p K
n 1/ n
1 1
\
z
H2
dH =
H1
z
z LMMNFGH IJK OPPQ
p2
Vdp
p1
H2 – H1 =
p1 V1n
p
p2
p1
1/ n
dp
FG p - p IJ
H 1-1/ n K
n bp V - p V g
=
1-1/ n
1
= (p1V n1)1/n
2 2
1 - 1/ n
1
1 1
n -1
=
a
f
1.3 3030.5 ¥ 0.05 - 500 ¥ 0.2
= 223.3 kJ
1.3 - 1
H2 – H1 = (U2 + p2V2) – (U1 + p1V1)
= (U2 – U1) + (p2 V2 – p1V1)
\
U2 – U1 = (H2 – H1) – (p2V2 – p1V1)
= 223.3 – 51.53 = 171.77 kJ
S2 – S1 = 0
Q1 – 2 = 0
Q1 – 2 = U2 – U 1 + W 1– 2
\
W1– 2 = U1 – U2 = – 171.77 kJ
Example 7.11 Air is flowing steadily in an insulated duct. The pressure and
temperature measurements of the air at two stations A and B are given below.
Establish the direction of the flow of air in the duct. Assume that for air, specific
v
0.287
heat cp is constant at 1.005 kJ/kg K, h = cp T, and
=
, where p, v, and T
p
T
are pressure (in kPa), volume (in m3/kg) and temperature (in K) respectively.
Pressure
Temperature
Solution
Station A
Station B
130 kPa
50°C
100 kPa
13°C
From property relation
Tds = dh – vdp
ds =
dh
dp
-v
T
T
196
Engineering Thermodynamics
For two states at A and B the entropy change of the system
z z
sB
sA
\
ds =
TB c dT
p
-
T
TA
z
pB
0.287
pA
dp
p
TB
p
– 0.287 ln B
TA
pA
sB – sA = 1.005 ln
100
273 + 13
– 0.287 ln
130
273 + 50
= 1.005 ln
= – 0.1223 + 0.0753
= – 0.047 kJ/kg K
(Ds)system = – 0.047 kJ/kg K
Since the duct is insulated (Ds)surr = 0
\
(Ds)univ = – 0.047 kJ/kg K
This is impossible. So the flow must be from B to A.
Example 7.12 A hypothetical device is supplied with 2 kg/s of air at 4 bar,
300 K. Two separate streams of air leave the device, as shown in figure below.
Each stream is at an ambient pressure of 1 bar, and the mass flow rate is the same
for both streams. One of the exit streams is said to be at 330 K while the other is at
270 K. The ambient temperature is at 300 K. Determine whether such a device is
possible.
Solution
The entropy generation rate for the control volume (Fig. 7.38) is
Air in
2 kg/s
4 bar, 300 K
Hot air
1 kg/s
Wx
Cold air
1 kg/s
Hypothetical device
1 bar, 270 K
1 bar, 330 K
C.V.
Q
Fig. 7.38
∑
∑
Sgen = S m e se - S m i si
∑
∑
= m 2 s2 + m 3 s3 - m 1 s1
∑
∑
∑
∑
= m 2 s2 + m 3 s3 - ( m 2 + m 3 )s1
∑
∑
∑
∑
= m 2 (s2 – s1) + m 3 (s3 – s1)
197
Entropy
Now,
s2 – s1 = cp ln
T2
p
– R ln 2
T1
p1
330
1
– 0.287 ln = 0.494 kJ/kg K
300
4
T3
p3
s3 – s1 = cp ln
– R ln
T1
p1
270
1
= 1.005 ln
– 0.287 ln = 0.292 kJ/kgK
300
4
= 1.005 ln
∑
Sgen = 1 ¥ 0.494 + 1 ¥ 0.292 = 0.786 kW/K
∑
Since Sgen > 0, the device is possible. Such devices actually exist and are called
vortex tubes. Although they have low efficiencies, they are suitable for certain
applications like rapid cooling of soldered parts, electronic component cooling,
cooling of machining operations and so on. The vortex tube is essentially a passive
device with no moving parts. It is relatively maintenance free and durable.
Example 7.13 A room is maintained at 27°C while the surroundings are at
2°C. The temperatures of the inner and outer surfaces of the wall (k = 0.71 W/mK)
are measured to be 21°C and 6°C, respectively. Heat flows steadily through the
wall 5 m ¥ 7 m in cross-section and 0.32 m in thickness. Determine (a) the rate of
heat transfer through the wall, (b) the rate of entropy generation in the wall, and
(c) the rate of total entropy generation with this heat transfer process.
Solution
∑
Q = kA
a
f
21 - 6 K
DT
W
= 0.71
¥ (5 ¥ 7) m2 ¥
mK
L
0.32m
= 1164.84 W
Taking the wall as the system, the entropy balance in rate form gives:
d Swall
= Stransfer + S gen.wall
dt
∑
∑
∑
0= S
0=
Q
∑
+ Sgen.wall
T
1164.84
-
1164.84
294
Rate of entropy generation in the wall
279
∑
∑
+ Sgen.wall
S gen.wall = 4.175 – 3.962 = 0.213 W/K
The entropy change of the wall is zero during this process, since the state and
hence the entropy of the wall does not change anywhere in the wall.
To determine the rate of total entropy generation during this heat transfer process,
we extend the system to include the regions on both sides of the wall,
d Stotal
= Stransfer + S gen.total
dt
∑
∑
198
Engineering Thermodynamics
∑
0= S
Q
∑
+ S gen.total
T
∑
1164.84 1164.84
0=
+ S gen.total
300
275
∑
S gen.total = 4.236 – 3.883 = 0.353 W/K
Summary
Two reversible adiabatic lines cannot intersect each other. So, a reversible
adiabatic line is a constant property sine. For a reversible cycle
dQ
ÑÚ T
=0
R
The entropy of a system is a quantity which satisfies the relation
dQ
R
R
where the heat dQ
is transferred reversibly.
ds =
Qrev = Ú Tds
If heat is supplied to a system, dQ
is positive, then ds is positive, and entropy
of the system increases. If heat is rejected from a system, dQ
is negative, and ds
is negative, and entropy of the system decreases. For a reversible adiabatic
is zero, and entropy remains constant, and the process is isentropic.
process, dQ
The inequality of Clausius states that the cyclic integral of
dQ
is negative for
T
an irreversible or actual cycle and zero for a reversible cycle, or Ñ
Ú
dQ
£ 0. If it
T
is positive, the cycle is impossible.
The entropy of an isolated system or universe always increases
dSiso ≥ 0
This is known as the principle of increase of entropy. The entropy of a system
and its surroundings together always increases in any process, and remains
constant only for a reversible process. The entropy increase of an isolated system
is a measure of the extent of irreversibility.
The entropy of a system can increase by external interaction (external
irreversibility) and internal irreversibility (like friction)
dS = deS + diS =
Since
dS >
dQ
,
T
dQ
d iS
T
\ d iS ≥ 0
Entropy
199
An isentropic process need not be reversible or adiabatic. If the isentropic
process is reversible, it must be adiabatic. If the isentropic process is adiabatic, it
cannot but be reversible. An adiabatic process need not be isentropic, since
entropy can also increase due to friction, etc.
By combining the first and second laws, the property relations are obtained
TdS = dU + pdV
TdS = dH – Vdp
All actual processes are irreversible. Between two equilibrium states a reversible
dQ
T
entropy increase for the actual process.
path is assumed and integration Ñ
Ú
is performed on that path to find the
Review Questions
7.1 Show that through one point there can pass only one reversible adiabatic.
7.2 State and prove Clausius’ theorem.
7.3 Show that entropy is a property of a system.
7.4 How is the entropy change of a reversible process estimated? Will it be different
for an irreversible process between the same end states?
7.5 Why is the Carnot cycle on T-s plot a rectangle?
7.6 Establish the inequality of Clausius.
7.7 Give the criteria of reversibility, irreversibility and impossibility of a thermodynamic cycle.
7.8 What do you understand by the entropy principle?
7.9 When the system is at equilibrium, why would any conceivable change in entropy be zero?
7.10 Why is the entropy increase of an isolated system a measure of the extent of
irreversibility of the process undergone by the system?
7.11 How did Rudolf Clausius summarize the first and second laws of thermodynamics?
7.12 Show that the transfer of heat through a finite temperature difference is irreversible.
7.13 Show that the adiabatic mixing of two fluids is irreversible.
7.14 What is the maximum work obtainable from two finite bodies at temperatures T 1
and T 2?
7.15 Determine the maximum work obtainable by using one finite body at temperature T and a thermal energy reservoir at temperature T 0, T > T0.
7.16 What are the causes of entropy increase?
7.17 Why is an isentropic process not necessarily an adiabatic process?
7.18 What is the reversible adiabatic work for a steady flow system when K.E. and
P.E. changes are negligibly small? How is it different from that for a closed stationary system?
200
Engineering Thermodynamics
7.19 Why are the equations
TdS = dU + pdV
TdS = dH – Vdp
valid for any process between two equilibrium end states?
7.20 Why is the second law called a directional law of nature?
7.21 How is entropy related to molecular disorder in a system?
7.22 Show that entropy varies logarithmically with the disorder number.
7.23 What do you understand by perfect order?
7.24 Give the Nernst-Simon statement of the third law of thermodynamics.
7.25 Why does entropy remain constant in a reversible adiabatic process?
7.26 What do you understand by the postulatory approach of thermodynamics?
7.27 What do you understand by ‘lost work’?
7.28 The amount of entropy generation quantifies the intrinsic irreversibility of a process. Explain.
7.29 Show that Sgen is not a thermodynamic property.
7.30 Give the expression for the entropy generation rate for a control volume of a
steady flow system.
7.31 What is the entropy generation in the isothermal dissipation of work?
7.32 What is the entropy generation in the adiabatic dissipation of work?
7.33 What do you understand by entropy transfer? Why is entropy transfer associated
with heat transfer and not with work transfer?
Problems
7.1 On the basis of the first law fill in the blank spaces in the following table of
imaginary heat engine cycles. On the basis of the second law classify each cycle
as reversible, irreversible, or impossible.
Cycle
Temperature
Rate of Heat Flow Rate of work Efficiency
Source Sink
Supply
Rejection
Output
(a) 327°C 27°C
420 kJ/s
230 kJ/s
... kW
(b) 1000°C 100°C
...kJ/min
4.2 MJ/min
...kW
65%
(c) 750 K 300 K
... kJ/s
... kJ/s
26 kW
60%
(d) 700 K 300 K 2500 kcal/h ... kcal/h
1 kW
—
Ans. (a) Irreversible (b) Irreversible, (c) Reversible, (d) Impossible
7.2 The latent heat of fusion of water at 0°C is 335 kJ/kg. How much does the
entropy of 1 kg of ice change as it melts into water in each of the following ways:
(a) Heat is supplied reversibly to a mixture of ice and water at 0°C. (b) A mixture
of ice and water at 0°C is stirred by a paddle wheel.
Ans. 1.2271 kJ/K
7.3 Two kg of water at 80°C are mixed adiabatically with 3 kg of water at 30°C in a
constant pressure process of 1 atmosphere. Find the increase in the entropy of
the total mass of water due to the mixing process (cp of water = 4.187 kJ/kg K).
Ans. 0.0576 kJ/K
7.4 In a Carnot cycle, heat is supplied at 350°C and rejected at 27°C. The working
fluid is water which, while receiving heat, evaporates from liquid at 350°C to
steam at 350°C. The associated entropy change is 1.44 kJ/kg K. (a) If the cycle
operates on a stationary mass of 1 kg of water, how much is the work done per
201
Entropy
p
p
cycle, and how much is the heat supplied? (b) If the cycle operates in steady flow
with a power output of 20 kW, what is the steam flow rate?
Ans. (a) 465.12, 897.12 kJ/kg, (b) 0.043 kg/s
7.5 A heat engine receives reversibly 420 kJ/cycle of heat from a source at 327°C,
and rejects heat reversibly to a sink at 27°C. There are no other heat transfers.
For each of the three hypothetical amounts of heat rejected, in (a), (b), and (c)
_
below, compute the cyclic integral of d Q/T. From these results show which case
is irreversible, which reversible, and which impossible: (a) 210 kJ/cycle rejected,
(b) 105 kJ/cycle rejected, (c) 315 kJ/cycle rejected.
Ans. (a) Reversible, (b) Impossible, (c) Irreversible
7.6 In Fig. 7.39, abcd represents a Carnot
cycle bounded by two reversible
a
adiabatics and two reversible isob
therms at temperatures T 1 and T 2 (T1
> T 2). The oval figure is a reversible
T1
d
cycle, where heat is absorbed at temperature less than, or equal to, T1, and
c T2
rejected at temperatures greater than,
or equal to, T 2. Prove that the effiv
ciency of the oval cycle is less than
Fig. 7.39
that of the Carnot cycle.
7.7 Water is heated at a constant pressure
of 0.7 MPa. The boiling point is 164.97°C. The initial temperature of water is
0°C. The latent heat of evaporation is 2066.3 kJ/kg. Find the increase of entropy
of water, if the final state is steam.
Ans. 6.6967 kJ/kg K
7.8 One kg of air initially at 0.7
MPa, 20°C changes to 0.35
Rev. isothermal
MPa, 60°C by the three reversa
1
ible non-flow processes, as
shown in Fig. 7.40. Process 10.35 MPa,
0.7 MPa,
a-2 consists of a constant presb
60ºC
20ºC
2
sure expansion followed by a
c
constant volume cooling, proRev. adiabatic
cess 1-b-2 an isothermal expansion followed by a constant
v
pressure expansion, and process 1-c-2 an adiabatic expanFig. 7.40
sion followed by a constant volume heating. Determine the change of internal energy, enthalpy, and entropy for
each process, and find the work transfer and heat transfer for each process. Take
cp = 1.005 and cv = 0.718 kJ/kg K and assume the specific heats to be constant.
Also assume for air pv = 0.287 T, where p is the pressure in kPa, v the specific
volume in m3/kg, and T the temperature in K.
202
Engineering Thermodynamics
7.9 Ten grammes of water at 20°C is converted into ice at –10°C at constant atmospheric pressure. Assuming the specific heat of liquid water to remain constant
at 4.2 J/gK and that of ice to be half of this value, and taking the latent heat
of fusion of ice at 0°C to be 335 J/g, calculate the total entropy change of the
system.
Ans. 16.02 J/K
7.10 Calculate the entropy change of the universe as a result of the following processes:
(a) A copper block of 600 g mass and with Cp of 150 J/K at 100°C is placed in
a lake at 8°C.
(b) The same block, at 8°C, is dropped from a height of 100 m into the lake.
(c) Two such blocks, at 100 and 0°C, are joined together.
Ans. (a) 6.69 J/K, (b) 2.095 J/K, (c) 3.64 J/K
7.11 A system maintained at constant volume is initially at temperature T 1, and a heat
reservoir at the lower temperature T 0 is available. Show that the maximum work
recoverable as the system is cooled to T0 is
W = Cv
LMbT - T g - T ln T OP
T Q
N
1
0
0
1
0
7.12 If the temperature of the atmosphere is 5°C on a winter day and if 1 kg of water
at 90°C is available, how much work can be obtained. Take cv of water as 4.186
kJ/kg K.
7.13 A body with the equation of state U = CT, where C is its heat capacity, is heated
from temperature T 1 to T2 by a series of reservoirs ranging from T 1 to T 2. The
body is then brought back to its initial state by contact with a single reservoir at
temperature T1. Calculate the changes of entropy of the body and of the reservoirs. What is the total change in entropy of the whole system?
If the initial heating were accomplished merely by bringing the body into contact
with a single reservoir at T 2, what would the various entropy changes be?
7.14 A body of finite mass is originally at temperature T 1 which is higher than that of
a reservoir at temperature T 2. Suppose an engine operates in a cycle between the
body and the reservoir until it lowers the temperature of the body from T1 to T2,
thus extracting heat Q from the body. If the engine does work W, then it will
reject heat Q-W to the reservoir at T2. Applying the entropy principle, prove that
the maximum work obtainable from the engine is
W (max) = Q – T 2 (S1 – S2)
where S1 – S2 is the entropy decrease of the body.
If the body is maintained at constant volume having constant volume heat
capacity Cv = 8.4 kJ/K which is independent of temperature, and if T1 = 373 K
and T 2 = 303 K, determine the maximum work obtainable.
Ans. 58.96 kJ
7.15 Each of three identical bodies satisfies the equation U = CT, where C is the heat
capacity of each of the bodies. Their initial temperatures are 200 K, 250 K, and
540 K. If C = 8.4 kJ/K, what is the maximum amount of work that can be extracted in a process in which these bodies are brought to a final common temperature?
Ans. 756 kJ
203
Entropy
7.16 In the temperature range between 0°C and 100°C a particular system maintained
at constant volume has a heat capacity.
CV = A + 2BT
with A = 0.014 J/K and B = 4.2 ¥ 10–4 J/K2.
A heat reservoir at 0°C and a reversible work source are available. What is the
maximum amount of work that can be transferred to the reversible work source
as the system is cooled from 100°C to the temperature of the reservoir?
Ans. 4.508 J
7.17 Each of the two bodies has a heat capacity at constant volume
CV = A + 2BT
where
A = 8.4 J/K and B = 2.1 ¥ 10–2 J/K2
If the bodies are initially at temperatures 200 K and 400 K and if a reversible
work source is available, what are the maximum and minimum final common
temperatures to which the two bodies can be brought? What is the maximum
amount of work that can be transferred to the reversible work source?
Ans. Tmin = 292 K
7.18 A reversible engine, as shown in Fig. 7.41 during a cycle of operations draws
5 MJ from the 400 K reservoir and does 840 kJ of work. Find the amount and
direction of heat interaction with other reservoirs.
200 K
300 K
Q3
Q2
400 K
Q1 = 5 MJ
E
W = 840 kJ
Fig. 7.41
Ans. Q2 = + 4.98 MJ
Q3 = – 0.82 MJ
7.19 For a fluid for which pv/T is a constant quantity equal to R, show that the change
in specific entropy between two states A and B is given by
sB – sA =
z FGH IJK
TB
Cp
TA
T
dT - R ln
pB
pA
A fluid for which R is a constant and equal to 0.287 kJ/kg K, flows steadily
through an adiabatic machine, entering and leaving through two adiabatic pipes.
In one of these pipes the pressure and temperature are 5 bar and 450 K and in the
other pipe the pressure and temperature are 1 bar and 300 K respectively. Determine which pressure and temperature refer to the inlet pipe.
Ans. A is the inlet pipe
204
Engineering Thermodynamics
For the given temperature range, cp is given by
cp = a ln T + b
where T is the numerical value of the absolute temperature and a = 0.026 kJ/kg
K, b = 0.86 kJ/kg K.
Ans. sB – sA = 0.0509 kJ/kg K
A is the inlet pipe.
7.20 Two vessels, A and B, each of volume 3 m3 may be connected by a tube of negligible volume. Vessel A contains air at 0.7 MPa, 95°C, while vessel B contains air
at 0.35 MPa, 205°C. Find the change of entropy when A is connected to B by
working from the first principles and assuming the mixing to be complete and
adiabatic. For air take the relations as given in Example 7.8.
Ans. 0.959 kJ/K
7.21 (a) An aluminium block (cp = 400 J/kg K) with a mass of 5 kg is initially at
40°C in room air at 20°C. It is cooled reversibly by transferring heat to a
completely reversible cyclic heat engine until the block reaches 20°C. The
20°C room air serves as a constant temperature sink for the engine. Compute (i) the change in entropy for the block, (ii) the change in entropy for the
room air, (iii) the work done by the engine.
(b) If the aluminium block is allowed to cool by natural convection to room air,
compute (i) the change in entropy for the block, (ii) the change in entropy
for the room air (iii) the net the change in entropy for the universe.
Ans. (a) – 134 J/K, + 134 J/K, 740 J,
(b) – 134 J/K, + 136.5 J/K, 2.5 J/K
7.22 Two bodies of equal heat capacities C and temperatures T1 and T 2 form an adiabatically closed system. What will the final temperature be if one lets this system
come to equilibrium (a) freely? (b) reversibly? (c) What is the maximum work
which can be obtained from this system?
7.23 A resistor of 30 ohms is maintained at a constant temperature of 27°C while a
current of 10 amperes is allowed to flow for 1 sec. Determine the entropy change
of the resistor and the universe.
Ans. (DS)resistor = 0, (DS)univ = 10 J/K
If the resistor initially at 27°C is now insulated and the same current is passed
for the same time, determine the entropy change of the resistor and the universe.
The specific heat of the resistor is 0.9 kJ/kg K and the mass of the resistor is 10 g.
Ans. (DS)univ = 6.72 J/K
7.24 An adiabatic vessel contains 2 kg of water at 25°C. By paddle-wheel work transfer, the temperature of water is increased to 30°C. If the specific heat of water is
assumed constant at 4.187 kJ/kg K, find the entropy change of the universe.
Ans. 0.139 kJ/K
7.25 A copper rod is of length 1 m and diameter 0.01 m. One end of the rod is at
100°C, and the other at 0°C. The rod is perfectly insulated along its length and
the thermal conductivity of copper is 380 W/mK. Calculate the rate of heat transfer along the rod and the rate of entropy production due to irreversibility of this
heat transfer.
Ans. 2.985 W, 0.00293 W/K
205
Entropy
7.26 A body of constant heat capacity Cp and at a temperature Ti is put in contact with
a reservoir at a higher temperature Tf . The pressure remains constant while the
body comes to equilibrium with the reservoir. Show that the entropy change of
the universe is equal to
Cp
LM T - T - ln F1 + T - T I OP
GH T JK P
MN T
Q
i
f
f
i
f
f
Prove that this entropy change is positive.
Given:
ln (1 + x) = x -
x2 x3 x4
... +
2
3
4
where x < 1.
7.27 An insulated 0.75 kg copper calorimeter can containing 0.2 kg water is in equilibrium at a temperature of 20°C. An experimenter now places 0.05 kg of ice at
0°C in the calorimeter and encloses the latter with a heat insulating shield. (a)
When all the ice has melted and equilibrium has been reached, what will be the
temperature of water and the can? The specific heat of copper is 0.418 kJ/kg K
and the latent heat of fusion of ice is 333 kJ/kg. (b) Compute the entropy increase of the universe resulting from the process. (c) What will be the minimum
work needed by a stirrer to bring back the temperature of water to 20°C?
Ans. (a) 4.68°C, (b) 0.00276 kJ/K, (c) 20.84 kJ
7.28 Show that if two bodies of thermal capacities C1 and C2 at temperatures T1 and
T 2 are brought to the same temperature T by means of a reversible heat engine,
then
C ln T1 + C2 ln T2
ln T = 1
C1 + C2
7.29 Two blocks of metal, each having a mass of 10 kg and a specific heat of
0.4 kJ/kg K, are at a temperature of 40°C. A reversible refrigerator receives
heat from one block and rejects heat to the other. Calculate the work required
to cause a temperature difference of 100°C between the two blocks.
7.30 A body of finite mass is originally at a temperature T 1, which is higher than that
of a heat reservoir at a temperature T 2. An engine operates in infinitesimal cycles
between the body and the reservoir until it lowers the temperature of the body
from T 1 to T2. In this process there is a heat flow Q out of the body. Prove that the
maximum work obtainable from the engine is Q + T2 (S1 – S2), where S1 – S2 is
the decrease in entropy of the body.
7.31 A block of iron weighing 100 kg and having a temperature of 100°C is immersed
in 50 kg of water at a temperature of 20°C. What will be the change of entropy of
the combined system of iron and water? Specific heats of iron and water are 0.45
and 4.18 kJ/kg K respectively.
Ans. 1.1328 kJ/K
7.32 36 g of water at 30°C are converted into steam at 250°C at constant atmospheric
pressure. The specific heat of water is assumed constant at 4.2 J/g K and the
latent heat of vaporization at 100°C is 2260 J/g. For water vapour, assume pV =
mRT where R = 0.4619 kJ/kg K, and
206
Engineering Thermodynamics
cp
= a + bT + cT 2, where a = 3.634,
R
b = 1.195 ¥ 10–3 K–1 and c = 0.135 ¥ 10–6 K–2
Calculate the entropy change of the system.
Ans. 277.8 J/K
7.33 A 50 ohm resistor carrying a constant current of 1 A is kept at a constant tem
perature of 27°C by a stream of cooling water. In a time interval of 1 s, (a) what
is the change in entropy of the resistor? (b) What is the change in entropy of the
universe?
Ans. (a) 0, (b) 0.167 J/K
7.34 A lump of ice with a mass of 1.5 kg at an initial temperature of 260 K melts at the
pressure of 1 bar as a result of heat transfer from the environment. After some
time has elapsed the resulting water attains the temperature of the environment,
293 K. Calculate the entropy production associated with this process. The latent
heat of fusion of ice is 333.4 kJ/kg, the specific heat of ice and water are 2.07
and 4.2 kJ/kg K respectively, and ice melts at 273.15 K.
Ans. 0.1514 kJ/K
7.35 An ideal gas is compressed reversibly and adiabatically from state a to state b. It
is then heated reversibly at constant volume to state c. After expanding reversibly and adiabatically to state d such that Tb = Td , the gas is again reversibly
heated at constant pressure to state e such that T e = T c. Heat is then rejected
reversibly from the gas at constant volume till it returns to state a. Express Ta in
terms of Tb and T c. If Tb = 555 K and T c = 835 K, estimate T a. Take g = 1.4.
Ans. T a =
Tbg + 1
, 313.29 K
Tcg
7.36 Liquid water of mass 10 kg and temperature 20°C is mixed with 2 kg of ice
at – 5°C till equilibrium is reached at 1 atm pressure. Find the entropy change of
the system. Given: cp of water = 4.18 kJ/kg K, cp of ice = 2.09 kJ/kg K and latent
heat of fusion of ice = 334 kJ/kg.
Ans. 190 J/K
7.37 A thermally insulated 50-ohm resistor carries a current of 1 A for 1 s. The initial
temperature of the resistor is 10°C. Its mass is 5 g and its specific heat is 0.85 J
g K. (a) What is the change in entropy of the resistor? (b) What is the change in
entropy of the universe?
Ans. (a) 0.173 J/K (b) 0.173 J/K
7.38 The value of cp for a certain substance can be represented by cp = a + bT. (a)
Determine the heat absorbed and the increase in entropy of a mass m of the
substance when its temperature is increased at constant pressure from T 1 to T2.
(b) Find the increase in the molal specific entropy of copper, when the temperature is increased at constant pressure from 500 to 1200 K. Given for copper:
when T = 500 K, cp = 25.2 ¥ 103 and when T = 1200 K, cp = 30.1 ¥ 103 J/k mol K.
LM b
MN
g b2 eT - T j, m LM a ln TT + b b T - T gOP OPP ,
QQ
N
Ans. (a) m a T2 - T1 +
2
2
2
1
2
2
2
1
(b) 24.7 kJ/k mol K
Entropy
207
7.39 Assume that a particular body has the equation of state U = CT, where C = 8.4 J/K
and assume that this equation of state is valid throughout the temperature range
from 0.5 K to room temperature. How much work must be expended to cool this
body from room temperature of 300 K to 0.5 K, taking the ambient atmosphere
as the hot reservoir?
7.40 An iron block of unknown mass at 85°C is dropped into an insulated tank that
contains 0.1 m3 of water at 20°C. At the same time a paddle-wheel driven by a
200 W motor is activated to stir the water. Thermal equilibrium is established
after 20 min when the final temperature is 24°C. Determine the mass of the iron
block and the entropy generated during the process.
Ans. 52.2 kg, 1.285 kJ/K
7.41 A piston cylinder device contains 1.2 kg of nitrogen gas at a 120 kPa and 27°C.
The gas is now compressed slowly in a polytropic process during which pV1.3 =
constant. The process ends when the volume is reduced by one-half. Determine
the entropy change of nitrogen during this process.
Ans. – 0.0615 kJ/K.
7.42 Air enters a compressor at ambient conditions of 96 kPa and 17°C with a low
velocity and exits at 1 MPa, 327°C, and 120 m/s. The compressor is cooled by
the ambient air at 17°C at a rate of 1500 kJ/min. The power input to the compressor is 300 kW. Determine (a) the mass flow rate of air and (b) the rate of entropy
generation.
Ans. (a) 0.851 kg/s, (b) 0.144 kW/K
7.43 A gearbox operating at steady state receives 0.1 kW along the input shaft and
delivers 0.095 kW along the output shaft. The outer surface of the gearbox is at
50°C. For the gearbox, determine (a) the rate of heat transfer, (b) the rate at
which entropy is produced.
Ans. (a) – 0.005 kW, (b) 1.54 ¥ 10–5 kW/K
7.44 At steady state, an electric motor develops power along its output shaft at the
rate of 2 kW while drawing 20 amperes at 120 volts. The outer surface of the
motor is at 50°C. For the motor, determine the rate of heat transfer and the rate of
entropy generation.
Ans. – 0.4 kW, 1.24 ¥ 10–3 kW/K
7.45 Show that the minimum theoretical work input required by a refrigeration cycle
to bring two finite bodies from the same initial temperature to the final temperatures of T 1 and T 2 (T 2 < T 1) is given by
Wmin = mc [2(T1T2)1/2 – T 1 – T 2]
7.46 A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel.
The paddle wheel does 200 kJ of work on the ideal gas. It is observed that the
temperature of the ideal gas remains constant during this process as a result of
heat transfer between the system and the surroundings at 25°C. Determine (a)
the entropy change of the ideal gas and (b) the total entropy generation.
Ans. (a) 0, (b) 0.671 kJ/K
7.47 A cylindrical rod of length L insulated on its lateral surface is initially in contact
at one end with a wall at temperature T 1 and at the other end with a wall at a
208
Engineering Thermodynamics
lower temperature T 2. The temperature within the rod initially varies linearly
with position x according to:
T1 - T2
x
L
The rod is insulated on its ends and eventually comes to a final equilibrium state
where the temperature is Tf . Evaluate Tf in terms of T 1 and T 2, and show that the
amount of entropy generated is:
T (x) = T 1 –
LM
N
Sgen = mc 1 + ln T f + +
T2
T1
ln T2 ln T1
T1 - T2
T1 - T2
OP
Q
where c is the specific heat of the rod.
Ans. T f = [T 1 + T 2]/2
7.48 Air flowing through a horizontal, insulated duct was studied by students in a
laboratory. One student group measured the pressure, temperature, and velocity
at a location in the duct as 0.95 bar, 67°C, 75 m/s. At another location the respective values were found to be 0.8 bar, 22°C, 310 m/s. The group neglected to note
the direction of flow, however. Using the known data, determine the
direction.
Ans. Flow is from right to left
7.49 Nitrogen gas at a 6 bar, 21°C enters an insulated control volume operating at
steady state for which WC.V. = 0. Half of the nitrogen exits the device at 1 bar,
82°C and the other half exits at 1 bar, – 40°C. The effects of KE and PE are
negligible. Employing the ideal gas model, decide whether the device can oper
ate as described.
Ans. Yes, the device can operate as described
7.50 An air compressor operates at steady state with air entering at p1 = 1 bar, T 1 =
20°C and exiting at p2 = 5 bar. Determine the work and heat transfer per unit
mass passing through the device, if the air undergoes a polytropic process with
n = 1.3. Neglect changes in KE and PE. Use air as an ideal gas.
Ans. – 164.2 kJ/kg, – 31 kJ/kg
8
Available Energy, Availability
and Irreversibility
8.1
AVAILABLE ENERGY
The sources of energy can be divided into two groups, viz. high grade energy and
low grade energy. The conversion of high grade energy to shaft work is exempt
from the limitations of the second law, while conversion of low grade energy is
subject to them.
The examples of two kinds of energy are:
(a)
(b)
(c)
(d)
(e)
(f)
High grade energy
Mechanical work
Electrical energy
Water power
Wind power
Kinetic energy of a jet
Tidal power
Low grade energy
(a) Heat or thermal energy
(b) Heat derived from nuclear
fission or fusion
(c) Heat derived from combustion
of fossil fuels
The bulk of the high grade energy in the form of mechanical work or electrical
energy is obtained from sources of low grade energy, such as fuels, through the
medium of the cyclic heat engine. The complete conversion of low grade energy,
heat, into high grade energy, shaft-work, is impossible by virtue of the second law
of thermodynamics. That part of the low grade energy which is available for
conversion is referred to as available energy, while the part which, according to the
second law, must be rejected, is known as unavailable energy.
Josiah Willard Gibbs is accredited with being the originator of the availability
concept. He indicated that environment plays an important part in evaluating the
available energy.
8.2
AVAILABLE ENERGY REFERRED TO A CYCLE
The maximum work output obtainable from a certain heat input in a cyclic heat
engine (Fig. 8.1) is called the available energy (A.E.), or the available part of the
energy supplied. The minimum energy that has to be rejected to the sink by the
210
Engineering Thermodynamics
second law is called the unavailable energy (U.E.), or the unavailable part of the
energy supplied.
Therefore,
Q1 = A.E. + U.E.
or
(8.1)
Wmax = A.E. = Q1 – U.E.
For the given T1 and T2,
h rev = 1 –
T2
T1
y
T1
dQ1
T
Q1
T1
Wmax = A.E.
E
x
Q2 = U.E.
T0
T2
Fig. 8.1
s
Available and
Unavailable Energy
in a Cycle
Fig. 8.2
Availability of Energy
For a given T1, h rev will increase with the decrease of T2. The lowest practicable
temperature of heat rejection is the temperature of the surroundings, T0.
\
h max = 1 –
and
Wmax = 1 -
FG
H
T0
T1
IJ
K
T0
Q1
T1
Let us consider a finite process x–y, in which heat is supplied reversibly to a heat
engine (Fig. 8.2). Taking an elementary cycle, if d-Q1 is the heat received by the
engine reversibly at T1, then
d-Wmax =
T1 - T0 T d Q1 = d-Q1 - 0 dQ
1 = A.E.
T1
T1
For the heat engine receiving heat for the whole process x-y, and rejecting heat
at T0
yy
yT
d Wmax = x d- Q1 - x 0 d-Q1
x
T1
\
Wmax = A.E. = Qxy – T0(sy – sx )
(8.2)
z
z
z
or
U.E. = Q xy – Wmax
or
U.E. = T0 (sy – sx )
Available Energy, Availability and Irreversibility
211
The unavailable energy is thus the product of the lowest temperature of heat
rejection, and the change of entropy of the system during the process of supplying
heat (Fig. 8.3). The available energy is also known as exergy and the unavailable
energy as anergy.
y
T
Qxy
Available energy,
Wmax = Wxy
x
T0
Unavailable
energy
s
Fig. 8.3
= T0 (sv – sx)
Unavailable Energy by the Second Law
8.2.1 Decrease in Available Energy when Heat is Transferred
through a Finite Temperature Difference
and
Q1
T1
T
Whenever heat is transferred through a finite
temperature difference, there is a decrease in
the availability of energy so transferred.
Let us consider a reversible heat engine
operating between T1 and T0 (Fig. 8.4). Then
Q1 = T1 Ds, Q2 = T0 Ds,
Wc
s
Fig. 8.4 Carnot Cycle
Q1
Q1
T
Let us now assume that heat Q1 is transferred through a finite temperature difference
from the reservoir or source at T1 to
the engine absorbing heat at T ¢1,
lower than T1 (Fig. 8.5). The availability of Q1 as received by the engine at T¢1 can be found by allowing
the engine to operate reversibly in a
cycle between T ¢1 and T0, receiving
Q1 and rejecting Q¢2.
Q1 = T1 Ds = T¢1 Ds¢
Since
T1 > T ¢1, \ Ds¢ > Ds
Since
D s¢ > D s
\ Q ¢2 > Q2
T1
T¢1
Q¢2
T0
Ds
Increase in
unavailable
energy
Ds¢
s
Q2 = T0 D s
Q¢2 = T0 D s¢
T0
Ds
W = A.E. = (T1 – T0) Ds
Now
WE
Q2
Fig. 8.5
Increase in Unavailable Energy Due
to Heat Transfer Through a Finite
Temperature Difference
212
Engineering Thermodynamics
\
W ¢ = Q1 – Q¢2 = T ¢1 Ds¢ – T0 Ds¢
and
W = Q1 – Q2 = T1 Ds – T0 Ds
\
W ¢ < W, because Q ¢2 > Q2
Available energy lost due to irreversible heat transfer through finite temperature
difference between the source and the working fluid during the heat addition process is given by
W – W¢ = Q¢2 – Q2
= T0 (Ds¢ – Ds)
or,
decrease in A.E. = T0 (Ds¢ – Ds)
The decrease in available energy or exergy is thus the product of the lowest
feasible temperature of heat rejection and the additional entropy change in the system while receiving heat irreversibly, compared to the case of reversible heat transfer from the same source.
The greater is the temperature difference (T1 – T ¢1 ), the greater is the heat rejection Q¢2 and the greater will be the unavailable part of the energy supplied or anergy
(Fig. 8.5). Energy is said to be degraded each time it flows through a finite temperature difference.
8.2.2
Available Energy from a Finite Energy Source
Let us consider a hot gas of mass mg at temperature T when the environmental
temperature is T0 (Fig. 8.6). Let the gas be cooled at constant pressure from state 1 at
temperature T to state 3 at temperature T0 and the heat given up by the gas, Q1, be
utilized in heating up reversibly a working fluid of mass mwf from state 3 to state 1
along the same path so that the temperature difference between the gas and the
working fluid at any instant is zero and hence, the entropy increase of the universe
is also zero. The working fluid expands reversibly and adiabatically in an engine or
turbine from sate 1 to state 2 doing work WE, and then rejects heat Q2 reversibly and
isothermally to return to the initial state 3 to complete a heat engine cycle.
T
p
1
mg
3
2
UE
5
Fig. 8.6
WE = AE
Q1
mwf
T0
Q2
v
4
Available Energy of a Finite Energy Source
Available Energy, Availability and Irreversibility
213
Here,
Q1 = mg cpg (T – T0) = mwf cpwf (T – T0)
= Area 14531
mg cpg = mwf cpwf
\
T
dT
= mg cpg ln 0 (negative)
T
T
dT
T
T
DSwf = T mwf c pwf
= mwf cpwf ln
(positive)
0
T
T0
z
T
DSgas = T 0 mg cpg
z
\
DSuniv = DSgas + DSwf = 0
Q2 = T0 DSwf = T0 mwf cpwf ln
\
T
T0
Available energy = Wmax
= Q1 – Q2
T
T0
= mg cpg (T – T0) – T0 mg cpg ln
= Area 1231
Therefore, the available energy or exergy of a gas of mass mg at temperature
T is given by
LM
N
OP
Q
T
T0
This is similar to Eq. (7.22) derived from the entropy principle.
AE = mg cpg ( T - T0 ) - T0 ln
(8.4)
8.3 QUALITY OF ENERGY
Let us assume that a hot gas is flowing through a pipeline (Fig. 8.7). Due to heat
loss to the surroundings, the temperature of the gas decreases continuously from
inlet at state a to the exit at state b. Although the process is irreversible, let us
assume a reversible isobaric path between the inlet and exit states of the gas
(Fig. 8.8). For an infinitesimal reversible process at constant pressure,
mc p dT
dS =
T
T
dT
or
=
(8.5)
dS
mcp
T ¢¢2
Tb
T ¢2
T ¢¢1
T ¢1
Ta
m
m
Q
T2
T1
Q
Fig. 8.7 Heat Loss From a Hot gas Flowing Through a Pipeline
214
Engineering Thermodynamics
where m is the mass of gas flowing and c p is its specific heat. The slope dT/dS
depends on the gas temperature T. As T increases, the slope increases, and if T
decreases the slope decreases.
a
T ¢1
T
T ¢¢1
1
T1
m
T ¢¢2
2
T ¢2
T2
Q
b
Q
T0
DS1
DS2
s
Fig. 8.8
Energy Quality at State 1 is Superior to that at State 2
Let us assume that Q units of heat are lost to the surroundings as the temperature
of the gas decreases from T ¢1 to T ¢¢1, T1 being the average of the two. Then,
Heat loss
Q = mcp (T ¢1 – T ¢¢1 ) = T1 DS1
(8.6)
Available energy lost with this heat loss at temperature T1 is
W1 = Q – T0 DS1
(8.7)
When the gas temperature has reached T2 (T2 < T1), let us assume that the same
heat loss Q occurs as the gas temperature decreases from T ¢2 to T ¢¢2, T2 being the
average temperature of the gas. Then
Heat loss Q = mcp (T ¢2 – T ¢¢2) = T2 DS2
(8.8)
Available energy lost with this heat loss at temperature T2 is
W2 = Q – T0 DS2
(8.9)
From Eqs. (8.6) and (8.8), since T1 > T2
DS1 < DS2
Therefore, from Eqs. (8.7) and (8.9),
W1 > W2
(8.10)
The loss of available energy is more, when heat loss occurs at a higher
temperature T1 than when the same heat loss occurs at a lower temperature T2.
Therefore, a heat loss of 1 kJ at, say, 1000°C is more harmful than the same heat
loss of 1 kJ at, say, 100°C. Adequate insulation must be provided for high
Available Energy, Availability and Irreversibility
215
temperature fluids (T >> T0) to prevent the precious heat loss. This may not be so
important for low temperature fluids (T ~ T0), since the loss of available energy
from such fluids would be low. (Similarly, insulation must be provided adequately
for very low temperature fluids (T << T0) to prevent heat gain from surroundings
and preserve available energy.)
The available energy or exergy of a fluid at a higher temperature T1 is more than
that at a lower temperature T2, and decreases as the temperature decreases.
The second law, therefore, affixes a quality to energy of a system at any state.
The quality of energy of a gas at, say, 1000°C is superior to that at, say, 100°C,
since the gas at 1000°C has the capacity of doing more work than the gas at 100°C,
under the same environmental conditions. An awareness of this energy quality as of
energy quantity is essential for the efficient use of our energy resources and for
energy conservation. The concept of available energy or exergy provides a useful
measure of this energy quality.
8.3.1
Law of Degradation of Energy
The available energy of a system decreases as its temperature or pressure decreases
and approaches that of the surroundings. When heat is transferred from a system,
its temperature decreases and hence the quality of its energy deteriorates. The degradation is more for energy loss at a higher temperature than that at a lower tem
perature. Quantitywise the energy loss may be the same, but qualitywise the losses
are different. While the first law states that energy is always conserved
quantitywise, the second law emphasizes that energy always degrades qualitywise.
When a gas is throttled adiabatically from a high to a low pressure, the enthalpy (or
energy per unit mass) remains the same, but there is a degradation of energy or
available work. The same holds good for pressure drop due to friction of a fluid
flowing through an insulated pipe. If the first law is the law of conservation of
energy, the second law is called the law of degradation of energy. Energy is
always conserved, but its quality is always degraded.
8.4 MAXIMUM WORK IN A REVERSIBLE PROCESS
Let us consider a closed stationary system undergoing a reversible process R from
state 1 to state 2 interacting with the surroundings at p0, T0 (Fig. 8.9). Then by the
first law,
QR = U2 – U1 + WR
(8.11)
If the process were irreversible, as represented by the dotted line I, connecting
the same equilibrium end states,
QI = U2 – U1 + WI
(8.12)
Therefore, from Eqs. (8.11) and (8.12),
QR – QI = WR – WI
Now,
DSsys = S2 – S1
(8.13)
216
Engineering Thermodynamics
Surroundings
p0,T0
1
System
1 Æ2
T
R
I
Q
W
2
s
Fig. 8.9
and
Maximum Work Done by a Closed System
DSsurr = -
Q
T0
By the second law,
DSuniv ≥ 0
For a reversible process,
DSuniv = S2 – S1 –
\
QR
=0
T0
QR = T0 (S2 – S1)
(8.14)
For an irreversible process,
DSuniv > 0
\
\
S2 – S1 –
QI
>0
T0
QI < T0 (S2 – S1)
From Eqs. (8.14) and (8.15),
QR > QI
(8.15)
(8.16)
Therefore, from Eqs. (8.13) and (8.16),
WR > WI
(8.17)
Therefore, the work done by a closed system by interacting only with the
surroundings at p0, T0 in a reversible process is always more than that done by it in
an irreversible process between the same end states.
8.4.1
Work Done in All Reversible Processes in the Same
Let us assume two reversible processes R1 and R2 between the same end states
1 and 2 undergone by a closed system by exchanging energy only with the surroundings (Fig. 8.10). Let one of the processes be reversed.
Available Energy, Availability and Irreversibility
217
Surroundings
p0,T0
1
System
1Æ 2
T
R1
W
Q
R2
2
s
Fig. 8.10 Equal Work Done in all Reversible Processes Between the Same End States
Then the system would execute a cycle 1–2–1 and produce net work represented
by the area enclosed by exchanging energy with only one reservoir, i.e. the surroundings. This violates the Kelvin-Planck statement. Therefore, the two
reversible processes must coincide and produce equal amounts of work.
8.5 REVERSIBLE WORK BY AN OPEN SYSTEM EXCHANGING
HEAT ONLY WITH THE SURROUNDINGS
Let us consider an open system exchanging energy only with the surroundings at
constant temperature T0 and at constant pressure p0 (Fig. 8.11). A mass dm1
enters the system at state 1, a mass dm2 leaves the system at state 2, an amount of
heat d- Q is absorbed by the system, an amount of work d- W is delivered by the
system, and the energy of the system (control volume) changes by an amount d(E)s .
Applying the first law, we have
FG
H
d- Q + dm1 h1 +
IJ
K
FG
H
IJ
K
V12
V2
+ gz1 – dm2 h2 + 2 + gz2 – d- W
2
2
LM
N
= dEs = d U +
OP
Q
mV 2
+ mgz c
2
s
(8.18)
T0
d Wmax = d W + d Wc
dQ0
d Wc
Surroundings
p0,T0
dm1(h1 +
V12
2
dW
+ gz1)
d(U +
1
E
dQ
T
dm2 h2 +
mV 2
+ mgz)s
2
C.V.
V22
2
+ gz2
2
C.S. s
Fig. 8.11
Reversible Work Done by an Open System While Exchanging Heat
Only With the Surroundings
218
Engineering Thermodynamics
For maximum work, the process must be entirely reversible. There is a temperature difference between the control volume and the surroundings. To make the heat
transfer process reversible, let us assume a reversible heat engine E operating between the two. Again, the temperature of the fluid in the control volume may be
different at different points. It is assumed that heat transfer occurs at points of the
control surface s where the temperature is T. Thus in an infinitesimal reversible
process an amount of heat d-Q0 is absorbed by the engine E from the surroundings
at temperature T0, an amount of heat d-Q is rejected by the engine reversibly to the
system where the temperature is T, and an amount of work d- Wc is done by the
engine. For a reversible engine,
dQ
d-Q
0
=
T
T0
\
d- Wc = d- Q0 – d- Q = d- Q
or
d- Wc = d-Q
T0
– d- Q
T
F T - 1I
HT K
0
(8.19)
The work d- Wc is always positive and is independent of the direction of heat
flow. When T0 > T, heat will flow from the surroundings to the system, d- Q is positive and hence d- Wc in Eq. (8.19) would be positive. Again, when T0 < T, heat will
flow from the system to the surroundings, d-Q is negative, and hence d-Wc would be
positive.
Now, since the process is reversible, the entropy change of the system will be
equal to the net entropy transfer, and Sgen = 0. Therefore,
dS =
Entropy
change
d-Q
+ dm1 s1 – dm2 s2
T
Entropy transfer
Entropy transfer with mass
with heat
d-Q
= dS – dm1s1 + dm1s2
(8.20)
T
Now, the maximum work is equal to the sum of the system work dW and the
work d- Wc of the reversible engine E,
\
d- Wmax = d- Wrev = d- W + d- Wc
(8.21)
From Eq. (8.19),
d- Wmax = d- W + d-Q
F T - 1I
HT K
0
(8.22)
Substituting Eq. (8.18) for d- W in Eq. (8.22),
FG
H
d- Wmax = d- Q + dm1 h1 +
IJ
K
FG
H
V12
V2
+ gz1 – dm2 h2 + 2 + gz2
2
2
IJ
K
Available Energy, Availability and Irreversibility
219
LM mV + mgz OP + dQ F T - 1I
HT K
N 2
Q
F V + gz IJ – dm FG h + V + gz IJ
= dm G h +
H 2 K
H 2 K
L mV + mgz OP + dQ T
– d MU +
(8.23)
N 2
Q T
2
0
-
–d U+
s
2
1
1
2
2
1
2
1
2
2
2
-
0
s
On substituting the value of d-Q/T from Eq. (8.20),
FG V + gz IJ – dm FG h + V + gz IJ
H 2 K
H 2 K
L mV + mgzOP + T (dS – dm s + dm s )
– d MU +
N 2
Q
F
I
F
I
V
V
= dm G h - T s +
+ gz J – dm G h - T s +
+ gz J
2
2
H
K
H
K
L
O
mV
– d MU - T S +
+ mgz P
(8.24)
2
N
Q
2
1
d- Wmax = dm1 h1 +
2
2
2
1
2
2
2
0
1 1
2 2
s
d- Wmax
\
2
1
1
1
2
2
0 1
2
1
2
0 2
2
2
0
s
Equation (8.24) is the general expression for the maximum work of an open
system which exchanges heat only with the surroundings at T0, p0.
8.5.1
Reversible Work in a Steady Flow Process
For a steady flow process
dm1 = dm2 = dm
LM
N
d U - T0 S +
and
OP
Q
mV 2
+ mgz = 0
2
s
Equation (8.24) reduces to
d- Wmax = dm
LMF h - T s + V + gz I - F h - T s + V + gz I OP
JK GH
JK PQ
2
2
MNGH
2
1
1
0 1
2
2
1
2
0 2
(8.25)
2
For total mass flow, the integral form of Eq. (8.25) becomes
FG
H
Wmax = H1 - T0 S1 +
IJ FG
K H
mV12
mV22
+ mgz1 - H2 - T0 S2 +
+ mgz2
2
2
IJ
K
The expression (H – T0S ) is called the Keenan function, B.
\
FG
H
Wmax = B1 +
IJ FG
K H
mV12
mV22
+ mgz1 - B2 +
+ mgz2
2
2
IJ
K
(8.26)
220
Engineering Thermodynamics
= y1 – y2
(8.27)
where y is called the availability function of a study flow process given by
y=B+
mV 2
+ mgz
2
On a unit mass basis,
FG
I F
I
V
V
+ gz J - G h - T s +
+ gz J
2
2
H
K H
K
F V + gz IJ - FG b + V + gz IJ
= Gb +
(8.28)
H 2 K H 2 K
Wmax = h1 - T0 s1 +
2
1
2
2
1
1
2
1
1
0 2
2
2
2
1
2
2
If K.E. and P.E. changes are neglected, Eqs. (8.27) and (8.28) reduce to
Wmax = B1 – B2
= (H1 – T0S1) – (H2 – T0S2)
= (H1 – H2) – T0(S1 – S2)
(8.29)
and per unit mass
Wmax = b1 – b2
= (h1 – h2) – T0 (s1 – s2)
8.5.2
(8.30)
Reversible Work in a Closed System
For a closed system,
dm1 = dm2 = 0
Equation (8.24) then becomes
LM
N
dWmax = – U - T0 S +
OP
Q
mV 2
+ mgz
2
s
= – d (E – T0 S )s
mV 2
+ mgz
2
For a change of state of the system from the initial state 1 to the final state 2,
where
E=U+
Wmax = E1 – E2 – T0 (S1 – S2)
= (E1 – T0S1) – (E2 – T0S2)
(8.31)
If the K.E. and P.E. changes are neglected, Eq. (8.31) reduces to
Wmax = (U1 – T0S1) – (U2 – T0S2)
(8.32)
For unit mass of fluid,
Wmax = (u1 – u2) – T0 (s1 – s2)
= (u1 – T0s1) – (u2 – T0s2)
(8.33)
Available Energy, Availability and Irreversibility
221
8.6 USEFUL WORK
All of the work W of the system would not be available for delivery, since a certain
portion of it would be spent in pushing out the atmosphere (Fig. 8.12). The useful
work is defined as the actual work delivered by a system less the work performed
on the atmosphere. If V1 and V2 are the initial and final volumes of the system
and p0 is the atmospheric pressure, then the work done on the atmosphere is
p0 (V2 – V1). Therefore, the useful work Wu becomes
Wu = Wact – p0 (V2 – V1)
(8.34)
V1
V2
System
Initial
Boundary
Final
Boundary
p0,T0
Surroundings
(Atmosphere)
Fig. 8.12
Work Done by a Closed System in Pushing Out the Atmosphere
Similarly, the maximum useful work will be
(Wu )max = Wmax – p0 (V2 – V1 )
(8.35)
In differential form
(dWu )max = dWmax – p0 dV
(8.36)
In a steady flow system, the volume of the system does not change. Hence,
the maximum useful work would remain the same, i.e. no work is done on the atmosphere, or
( d- Wu )max = d- Wmax
(8.37)
But in the case of an unsteady-flow open system or a closed system, the volume of
the system changes. Hence, when a system exchanges heat only with the atmosphere, the maximum useful work becomes
( d- Wu ) max = d- Wmax – p0 dV
Substituting dWmax from Eq. (8.24),
F
I
F
I
V
V
+ gz J – dm G h - T s +
+ gz J
GH
2
2
K
H
K
L
O
mV
– d MU + p V - T S +
+ mgz P
2
N
Q
2
1
( d- Wu ) = dm1 h1 - T0 s1 +
2
2
1
2
2
0 2
2
2
0
0
s
(8.38)
222
Engineering Thermodynamics
This is the maximum useful work for an unsteady open system. For the closed
system Eq. (8.38) reduces to
LM
N
( d-Wu ) max = –d U + p0 V - T0 S +
OP
Q
mV 2
+ mgz
2
s
= – d[E + p0 V – T0 S ]s
\
(8.39)
(Wu)max = E1 – E2 + p0 (V1 – V2) – T0 (S1 – S2)
(8.40)
If K.E. and P.E. changes are neglected, Eq. (8.40) becomes
(Wu )max = U1 – U2 + p0 (V1 – V2) – T0 (S1 – S2)
(8.41)
This can also be written in the following form
(Wu )max = (U1 + p0V1 – T0S1) – (U2 + p0V2 – T0S2)
(8.42)
= f1 – f2
where f is called the availability function for a closed system given by
f = U + p0V – T0S
The useful work per unit mass becomes
(Wu ) max = (u1 + p0v1 – T0s1) – (u2 + p0v2 – T0s2)
(8.43)
8.6.1 Maximum Useful Work Obtainable when the System
Exchanges Heat with a Thermal Reservoir in Addition to
the Atmosphere
If the open system discussed in Sec. 8.5 exchanges heat with a thermal energy reservoir at temperature TR in addition to the atmosphere, the maximum useful work
FG
H
will be increased by d- QR 1 -
IJ
K
T0
, where d- QR is the heat received by the system.
TR
For a steady flow process,
F
GH
(Wu)max = Wmax = H1 - T0 S1 +
F
GH
– H2 - T0 S2 +
mV12
+ mgz1
2
mV22
+ mgz2
2
FG
H
I
JK
I + Q F1 - T I
JK GH T JK
0
R
R
IJ
K
= y1 – y2 + QR 1 -
T0
TR
FG
H
T0
TR
(8.44)
For a closed system
(Wu )max = Wmax – p0 (V2 – V1) + QR 1 -
or
IJ
K
FG
H
(Wu )max = E1 – E2 + p0 (V1 – V2 ) – T0 (S1 – S2 ) + QR 1 -
T0
TR
IJ
K
(8.45)
Available Energy, Availability and Irreversibility
223
If K.E. and P.E. changes are neglected, then for a steady flow process:
FG
H
T0
TR
IJ
K
(8.46)
FG
H
T0
TR
IJ
K
(8.47)
(Wu )max = (H1 – H2) – T0 (S1 – S2) + QR 1 and for a closed system:
(Wu) max = U1 + U2 – p0 (V1 – V2) – T0 (S1 – S2) + QR 1 -
8.7 DEAD STATE
If the state of a system departs from that of the surroundings, an opportunity exists
for producing work (Fig. 8.13). However, as the system changes its state towards
that of the surroundings, this opportunity diminishes, and it ceases to exist when the
two are in equilibrium with each other. When the system is in equilibrium with the
surroundings, it must be in pressure and temperature equilibrium with the surroundings, i.e. at p0 and T0. It must also be in chemical equilibrium with the surroundings,
i.e. there should not be any chemical reaction or mass transfer. The system must
have zero velocity and minimum potential energy. This state of the system is known
as the dead state, which is designated by affixing subscript ‘0’ to the properties.
Any change in the state of the system from the dead state is a measure of the available work that can be extracted from it. Farther the initial point of the system from
the dead state in terms of p, t either above or below it, higher will be the available
energy or exergy of the system (Fig. 8.13). All spontaneous processes terminate
at the dead state.
1
p0
1
p
Availability
0
p0
Dead
State
Availability
Isotherm
at T0
1¢
T
0
Dead State
T0
1
s
v
(a)
(b)
Fig. 8.13 Available Work of a System Decreases as its State Approaches P0, T0
8.8
AVAILABILITY
Whenever useful work is obtained during a process in which a finite system
undergoes a change of state, the process must terminate when the pressure and
temperature of the system have become equal to the pressure and temperature of
the surroundings, p0 and T0, i.e. when the system has reached the dead state. An air
224
Engineering Thermodynamics
engine operating with compressed air taken from a cylinder will continue to deliver
work till the pressure of air in the cylinder becomes equal to that of the surroundings, p0. A certain quantity of exhaust gases from an internal combustion engine
used as the high temperature source of a heat engine will deliver work until the
temperature of the gas becomes equal to that of the surroundings, T0.
The availability (A) of a given system is defined as the maximum useful work
(total work minus pdV work) that is obtainable in a process in which the system
comes to equilibrium with its surroundings. Availability is thus a composite property depending on the state of both the system and surroundings.
8.8.1
Availability in a Steady Flow Process
The reversible (maximum) work associated with a steady flow process for a single
flow is given by Eq. (8.26)
FG
H
Wrev = H1 - T0 S1 +
IJ FG
K H
mV12
mV22
+ mgz1 - H2 - T0 S2 +
+ mgz2
2
2
IJ
K
With a given state for the mass entering the control volume, the maximum useful
work obtainable (i.e., the availability) would be when this mass leaves the control
volume in equilibrium with the surroundings (i.e. at the dead state). Since there is
no change in volume, no work will be done on the atmosphere. Let us designate the
initial state of the mass entering the C.V. with parameters having no subscript and
the final dead state of the mass leaving the C.V. with parameters having subscript 0.
The maximum work or availability, A, would be
FG
H
A = H - T0 S +
IJ
K
mV 2
+ mgz – (H0 – T0S0 + mgz0) = y – y0
2
(8.48)
where y is called the availability function for a steady flow system and V0 = 0. This
is the availability of a system at any state as it enters a C.V. in a steady flow process.
The availability per unit mass would be
FG
H
a = h - T0 s +
IJ
K
V2
+ gz – (h0 – T0s0 + gz) = y – y0
2
(8.49)
If subscripts 1 and 2 denote the states of a system entering and leaving a C.V. the
decrease in availability or maximum work obtainable for the given system-surroundings combination would be
Wmax = a1 – a2 = y1 – y2
=
LMF h - T s + V + gz I - b h - T s + gz gOP
JK
2
MNGH
PQ
LF
O
I
V
– MG h - T s +
+ gz J - b h - T s + gz g P
2
K
MNH
PQ
2
1
1
0 1
1
0
0 0
0
2
2
2
0 0
2
= (h1 – h2) – T0 (s1 – s2) +
0
0 0
0
V12 - V22
+ g (z1 – z2)
2
(8.50)
Available Energy, Availability and Irreversibility
225
If K.E. and P.E. changes are neglected,
Wmax = (h1 – T0s1) – (h2 – T0s2)
= b1 – b2
where b is the specific Keenan function.
If more than one flow into and out of the C.V. is involved.
Wmax =  mi yi –  meye
i
8.8.2
e
Availability in a Nonflow Process
Let us consider a closed system and denote its initial state by parameters without
any subscript and the final dead state with subscript ‘0’. The availability of the
system A, i.e. the maximum useful work obtainable as the system reaches the dead
state, is given by Eq. (8.40)
A = (Wu )max = E – E0 + p0 (V – V0) – T0(S – S0)
FG
H
= U+
IJ
K
mV 2
+ mgz – (U0 + mgz0) + p0 (V – V0) – T0 (S – S0)
2
(8.51)
If K.E. and P.E. changes are neglected and for unit mass, the availability
becomes
a = u – u0 + p0 (v – v0) – T0 (s – s0)
= (u + p0v – T0s) – (u0 + p0v0 – T0s0)
= f – f0
(8.52)
where f is the availability function of the closed system.
If the system undergoes a change of state from 1 to 2, the decrease in availability
will be
a = (f1 – f 0) – (f 2 – f0)
= f1 – f 2
= (u1 – u2) + p0 (v1 – v2) – T0 (s1 – s2)
(8.53)
This is the maximum useful work obtainable under the given surroundings.
8.9 AVAILABILITY IN CHEMICAL REACTIONS
In many chemical reactions the reactants are often in pressure and temperature equilibrium with the surroundings (before the reaction takes place) and so are the products after the reaction. An internal combustion engine can be cited as an example of
such a process if we visualize the products being cooled to atmospheric temperature T0 before being discharged from the engine.
(a) Let us first consider a system which is in temperature equilibrium with the
surroundings before and after the process. The maximum work obtainable during a
change of state is given by Eq. (8.31),
Wmax = E1 – E2 – T0 (S1 – S2)
226
Engineering Thermodynamics
F
GH
= U1 +
I F
JK GH
I
JK
mV12
mV22
+ mgz1 - U2 +
+ mgz2 – T0 (S1 – S2)
2
2
If K.E. and P.E. changes are neglected,
Wmax = U1 – U2 – T0 (S1 – S2)
Since the initial and final temperatures of the system are the same as that of the
surroundings, T1 = T2 = T0 = T, say, then
(WT )max = (U1 – U2)T – T (S1 – S2)T
(8.54)
Let a property called Helmholtz function F be defined by the relation
F = U – TS
(8.55)
Then for two equilibrium states 1 and 2 at the same temperature T,
(F1 – F2)T = (U1 – U2)T – T (S1 – S2)
From Eqs. (8.54) and (8.56),
(WT )max = (F1 – F2 )T
(8.56)
(8.57)
WT £ (F1 – F2)T
(8.58)
The work done by a system in any process between two equilibrium states at the
same temperature during which the system exchanges heat only with the environment is equal to or less than the decrease in the Helmholtz function of the system
during the process. The maximum work is done when the process is reversible and
the equality sign holds. If the process is irreversible, the work is less than the maximum.
(b) Let us now consider a system which is in both pressure and temperature
equilibrium with the surroundings before and after the process. When the volume
of the system increases, some work is done by the system against the surroundings
( pdV work), and this is not available for doing useful work. The availability of the
system, as defined by Eq. (8.51), neglecting the K.E. and P.E. changes, can be
expressed in the form
A = (Wu )max = (U + p0V – T0S) – (U0 + p0V0 – T0S0)
or
= f – f0
The maximum work obtainable during a change of state is the decrease in availability of the system, as given by Eq. (8.53) for unit mass.
\
(Wu )max = A1 – A2 = f1 – f2
= (U1 – U2) + p0(V1 – V2) – T0(S1 – S2)
If the initial and final equilibrium states of the system are at the same pressure
and temperature of the surroundings, say p1 = p2 = p0 = p, and T1 = T2 = T0 = T.
Then,
(Wu )max = (U1 – U2) p,T + p (V1 – V2) p,T – T (S1 – S2)p,T
(8.59)
The Gibbs function G is defined as
G = H – TS
= U + pV – TS
(8.60)
Available Energy, Availability and Irreversibility
227
Then for two equilibrium states at the same pressure p and temperature T
(G1 – G2) p,T = (U1 – U2)p,T + p(V1 – V2)p,T – T (S1 – S2) p,T
(8.61)
From Eqs. (8.59) and (8.61),
(Wu ) max = (G1 – G2)p,T
(8.62)
p,T
\
(W u ) p,T £ (G1 – G2) p,T
(8.63)
The decrease in the Gibbs functions of a system sets an upper limit to the work that
can be performed, exclusive of pdV work, in any process between two equilibrium
states at the same temperature and pressure, provided the system exchanges heat
only with the environment which is at the same temperature and pressure as the end
states of the system. If the process is irreversible, the useful work is less than the
maximum.
8.10 IRREVERSIBILITY AND GOUY-STODOLA THEOREM
The actual work done by a system is always less than the idealized reversible work,
and the difference between the two is called the irreversibility of the process.
I = Wmax – W
(8.64)
This is also sometimes referred to as ‘degradation’ or ‘dissipation’.
For a non-flow process between the equilibrium states, when the system
exchanges heat only with the environment
I = [(U1 – U2) – T0 (S1 – S2)] – [(U1 – U2) + Q]
= T0 (S2 – S1) – Q
= T0 (DS)system + T0 (DS)surr
= T0 [(DS)system + (DS)surr] = T0(DS)univ.
\
I≥0
Similarly, for the steady flow process
I = Wmax – W
(8.65)
éæ
ö æ
öù
mV 2
mV22
+ mgZ 2 ÷ ú
= êç B1 + 1 + mgZ1 ÷ - ç B2 +
2
2
ø è
ø ûú
ëêè
–
LMF H + mV + mgZ I - F H + mV + mgZ I + QOP
JK GH
JK PQ
2
2
MNGH
2
1
1
2
2
1
2
2
= T0(S2 – S1) – Q
= T0(DS)system + T0 (DS)surr
= T0(DSsystem DSsurr ) = T0 DSuniv
(8.66)
The same expression for irreversibility applies to both flow and non-flow
processes. The quantity T0 (DSsystem + DSsurr ) represents an increase in unavailable
energy (or anergy).
228
Engineering Thermodynamics
The Gouy-Stodola theorem states that the rate of loss of available energy or
.
exergy in a process is proportional to the rate of entropy generation, Sgen . If
Eqs. (8.65) and (8.66) are written in the rate form,
.
.
.
.
I = Wlost = T0 DSuniv = T0 Sgen
(8.67)
This is known as the Gouy-Stodola equation. A thermodynamically efficient process would involve minimum exergy loss with minimum rate of entropy generation.
8.10.1
Application of Gouy-Stodola Equation
(a) Heat Transfer through a Finite Temperature Difference If heat
.
transfer Q occurs from the hot reservoir at temperature T1 to the cold reservoir at
temperature T2 (Fig. 8.14a)
T1
T1
Q
Q
E
Brake
W
Q –W
W
Q
T2
T2
(a)
(b)
Fig. 8.14
Destruction of Available Work or Exergy by Heat Transfer
Through a Finite Temperature Difference
.
.
. T -T
Q Q
Sgen =
=Q 1 2
T1T2
T2 T1
and
\
FG
H
.
.
T
Wlost = Q 1 - 2
T1
.
Wlost = T2 Sgen
IJ = Q. T - T
T
K
1
2
1
.
If the heat transfer.Q from T1 to T2 takes place through a reversible engine E, the
entire work output W is dissipated in the brake, from which an equal amount of
heat is rejected to the reservoir at T2 (Fig. 8.14b). Heat transfer through a finite
temperature difference is equivalent to the destruction of its availability.
(b) Flow with Friction
Let us consider the steady and adiabatic flow of an ideal
gas through the segment of a pipe (Fig. 8.15a).
By the first law,
h1 = h2
and by the second law,
Tds = dh – vdp
Available Energy, Availability and Irreversibility
Insulation
1
229
2
B1
B2
(mb) out
(mb) in
p1
p2
Dp = p1 – p2
WLOST
(a)
(b)
Irreversibility in a Duct Due to Fluid Friction
Fig. 8.15
z ds = z dhT - z uT dp = - z uT dp
2
1
2
1
2
1
2
1
∑
z
2 ∑
z
p2 ∑
Sgen = 1 mds = - p1 mR
\
FG
H
∑
= - m R ln 1 ∑
= + mR
where
FG
H
ln 1 -
Dp
p1
p
dp
= - m R ln 2
p
p1
IJ
K
∑
FG
H
Dp
Dp
= - mR p1
p1
∑
Dp
p1
IJ
K
(8.68)
IJ ~= - D p , since D p < 1
p
K p
1
1
and higher terms are neglected.
.
.
∑
Wlost = B 1 - B 2
∑
= m [(h1 – T0s1) – (h2 – T0s2)]
∑
= mT0 (s2 – s1 )
Dp
.
= T0 Sgen = m RT0
(8.69)
p1
The decrease in availability or lost work is proportional to the pressure drop (D p)
.
and the mass flow rate ( m ) . It is shown on the right (Fig. 8.15b) by the Grassmann
diagram, the width being proportional to the availability (or exergy) of the stream.
It is an adaptation of the Sankey diagram used for energy transfer in a plant.
∑
(c) Mixing of Two Fluids Two streams 1 and 2 of an incompressible fluid or an
ideal gas mix adiabatically at constant pressure (Fig. 8.16).
.
.
Here,
m1 + m 2 = m 3 = m (say)
∑
∑
∑
Let
m1
x=
∑
∑
m1 + m 2
230
Engineering Thermodynamics
m1
(mb)1
(mb)2
m3
3
m2
WLost
Insulation
(b)
(a)
Fig. 8.16
Irreversibility Due to Mixing
By the first law,
∑
∑
∑
∑
m1 h1 + m 2 h2 = ( m1 + m 2 ) h3
or
xh1 + (1 – x) h2 = h3
The preceding equation may be written in the following form, since enthalpy is a
function of temperature.
xT1 + (1 – x)T2 = T3
T3
\
T1
= x + (1 – x) t
t=
where
(8.70)
T2
T1
By the second law,
∑
∑
∑
∑
∑
∑
S gen = m 3 s3 - m1 s1 - m 2 s2
∑
= ms3 - x ms1 – (1 – x) ms2
∑
\
Sgen
= (s3 – s2) + x (s2 – s 1)
∑
m
= cp ln
T3
T2
+ xcp ln
FT I FT I
= ln G J G J
HT K HT K
mc
FT I T
N = ln G J
HT K T
∑
or
S gen
∑
p
or
3
2
2
1
3
s
1
or
Ns = ln
T2
T1
x
1- x
1
1- x
2
T3 / T1
(T2 / T1 )1 - x
(8.71)
Available Energy, Availability and Irreversibility
231
where NS is a dimensionless quantity, called the entropy generation number, given
∑
∑
by S gen / mc p .
Substituting T3/T1 from Eq. (8.70) in Eq. (8.71),
x + t (1 - x)
Ns = ln
(8.72)
t1- x
If x = 1 or t = 1, Ns becomes zero in each case. The magnitude of Ns depends on
x and t. The rate of loss of exergy due to mixing would be
x + t (1 - x)
∑
W lost = I = T0 mc p ln
(8.73)
t1- x
∑
∑
8.11 AVAILABILITY OR EXERGY BALANCE
The availability or exergy is the maximum useful work obtainable from a system as it
reaches the dead state (p0, t 0). Conversely, availability or exergy can be regarded as
the minimum work required to bring the closed system from the dead state to the
given state. The value of exergy cannot be negative. If a closed system were at any
state other than the dead state, the system would be able to change its state spontaneously toward the dead state. This tendency would stop when the dead state is
reached. No work is done to effect such a spontaneous change. Since any change in
state of the closed system to the dead state can be accomplished with zero work, the
maximum work (or exergy) cannot be negative.
While energy is always conserved, exergy is not generally conserved, but is
destroyed by irreversibilities. When the closed system is allowed to undergo a
spontaneous change from the given state to the dead state, its exergy is completely
destroyed without producing any useful work. The potential to develop work that
exists originally at the given state is thus completely wasted in such a spontaneous
process. Therefore, at steady state:
1. Energy in – Energy out = 0
2. Exergy in – Exergy out = Exergy destroyed
8.11.1
Exergy Balance for a Closed System
For a closed system, availability or exergy transfer occurs through heat and work
interactions (Fig. 8.17).
Q
Ts
1
Boundary
Fig. 8.17
2
W1 – 2
Exergy Balance For a Closed System
232
Engineering Thermodynamics
2
z
E2 – E1 = d-Q – W 1– 2
1st law:
(8.74)
1
L d Q OP = S
2nd law:
S –S – M
MN T PQ
L d Q OP = T S
or
T (S – S ) – T M
MN T PQ
-
2
2
z
1
gen
1
-
2
0
2
1
z
0
0
1
(8.75)
gen
s
Subtracting Eq. (8.75) from Eq. (8.74),
z
z LMMN
1
2
=
1-
1
LM d Q OP – T S
MN T PQ
-
2
2
E2 – E1 – T0 (S2 – S1) = d-Q – W 1– 2 – T0
z
0
1
gen
OP d Q - W
T PQ
T0
-
1 - 2 – T0 S gen
s
Since,
A2 – A1 = E2 – E1 + p0(V2 – V1) – T0(S2 – S1)
z LMMN
2
A2 – A1 =
1-
1
Change
in exergy
OP d Q – [W
T PQ
T0
-
1 – 2 – p0(V2 – V1)] – T0 S gen
(8.76)
s
Exergy transfer
with heat
Exergy transfer
with work
Exergy
destruction
In the form of the rate equation,
dA
dt
Rate of
change of
exergy
LM T OP Q
MN T PQ
Rate of exergy
= S 1j
0
∑
j
j
transfer with heat
Q j at the boundary
where the instantaneous
temperature is Tj
LM
N
- W - p0
OP
Q
dV
dt
Rate of exergy
transfer as work
where dV/dt is
the rate of
change of system
volume
∑
I
(8.77)
Rate of exergy
loss due to
irreversibilities
◊
( = T0 Sgen )
For an isolated system, the exergy balance, Eq. (8.77), gives
DA = – I
(8.78)
Since I > 0, the only processes allowed by the second law are those for which the
exergy of the isolated system decreases. In other words,
The exergy of an isolated system can never increase.
It is the counterpart of the entropy principle which states that the entropy of an
isolated system can never decrease.
The exergy balance of a system can be used to determine the locations, types and
magnitudes of losses (waste) of the potential of energy resources (fuels) and find
ways and means to reduce these losses of making the system more energy efficient
and for more effective use of fuel.
Available Energy, Availability and Irreversibility
8.11.2
233
Exergy Balance for a Steady Flow System
1st law:
H1 +
= H2 +
m V12
2
m V22
2
2
2nd law:
S1 +
+ mgZ1 + Q1– 2
+ mgZ2 + W1 – 2
(8.78)
LM d Q OP – S = S
NTQ
LM d Q OP = T S = I
MN T PQ
-
z
2
gen
1
2
or
z
T0 (S1 – S2) + T0
1
W1 – 2
-
0
(8.79)
gen
s
C.S.
1
2
C.V.
m1
m2
Ts
Q1 – 2
Exergy Balance for a Steady Flow System
Fig. 8.18
From Eqs. (8.78) and (8.79),
m (V22 - V12 )
+ mg (Z2 – Z1)
2
H2 – H1 – T0 (S 2 – S1) +
z FGH
F
z GH
2
=
1
2
or
A2 – A1 =
1
I dQ – W
T JK
T I
1 - J dQ – W
T K
1-
T0
-
1–2 – I
(8.80)
-
1– 2 – I
(8.81)
s
0
s
In the form of rate equation at steady state:
F
GH
 1j
I Q - W + m (a - a ) - I
T JK
T0
∑
∑
j
2
f1
j
where af – a f = (h1 – h2) – T0 (s1 – s2) +
1
∑
∑
C.V
V12 - V22
f2
C.V. = 0
(8.82)
◊
+ g (Z1 – Z2) and [1 – T0/Tj] Q j
2
∑
= time rate of exergy transfer along with heat Q j occurring at the location on the
boundary where the instantaneous temperature is Tj.
234
Engineering Thermodynamics
For a single stream entering and leaving, the exergy balance gives
∑
LM1 - T OP Q + a - W - a = I
m
m
MN T PQ m
∑
∑
0
∑
s
Exergy in
8.12
f1
∑
f2
∑
(8.83)
Exergy out Exergy loss
SECOND LAW EFFICIENCY
A common measure on energy use efficiency is the first law efficiency, hI . The first
law efficiency is defined as the ratio of the output energy of a device to the input
energy of the device. The first law is concerned only with the quantities of energy,
and disregards the forms in which the energy exists. It does not also discriminate
between the energies available at different temperatures. It is the second law of
thermodynamics which provides a means of assigning a quality index to energy. The
concept of available energy or exergy provides a useful measure of energy quality
(Sec. 8.3).
With this concept it is possible to analyze means of minimizing the consumption
of available energy to perform a given process, thereby ensuring the most efficient
possible conversion of energy for the required task.
The second law efficiency, h II , of a process is defined as the ratio of the minimum
available energy (or exergy) which must be consumed to do a task divided by the
actual amount of available energy (or exergy) consumed in performing the task.
minimum exergy intake to perform the given task
hII =
actual exergy intake to perform the same task
or
hII =
A min
(8.84)
A
where A is the availability or exergy.
A power plant converts a fraction of available energy A or W max to useful work W.
For the desired output of W, Amin = W and A = Wmax . Here,
W
I = Wmax – W and hII =
(8.85)
Wmax
Now
W
W Wmax
hI =
=
◊
Q1 Wmax Q1
= h I I ◊ h Carnot
hI
h II =
h Carnot
\
FG
H
Since Wmax = Q 1 1 -
T0
T
(8.86)
(8.87)
IJ , Eq. (8.87) can also be obtained directly as follows
K
h II =
W
=
T I h
F
Q G1 - J
H TK
0
1
hI
Carnot
235
Available Energy, Availability and Irreversibility
FG
H
If work is involved, Amin = W (desired) and if heat is involved, Amin = Q 1 -
T0
IJ
K
.
T
If solar energy, Q r is available at a reservoir storage temperature Tr and if quantity
of heat Q a is transferred by the solar collector at temperature Ta, then
hI =
h II =
and
Qa
Qr
exergy output
exergy input
F TI
GH T JK
=
F TI
Q G1 - J
H TK
Qa 1 -
0
a
0
r
r
T0
Ta
= hI
T
1- 0
Tr
1-
(8.88)
Table 8.1 shows availabilities, and both hI and hII expressions for several common thermal tasks.
Table 8.1
Tr > Ta > T0 > Tc
Task
Produce work, W 0
Energy input
Input shaft work, Wi
Q r from reservoir at T r
A = Wi
A = Q r Á1 -
Amin = W 0
Amin = W 0
hI =
h II =
W0
Wi
Amin
A
hI =
Ê
T0 ˆ
Ë
Tr ¯
˜
W0
Qr
h II = h I ◊
1
T0
1Tr
h II = h I
Add heat Qa to
reservoir at Ta
(electric motor)
(heat engine)
A = Wi
A = Qr 1 -
F
GH
I
J
T K
T0
r
(Contd.)
236
Table 8.1
Engineering Thermodynamics
(Contd.)
Task
Energy input
Input shaft work, Wi
F
GH
Amin = Q a 1 -
Q r from reservoir at T r
I
J
T K
F
GH
T0
Amin = Q a 1 -
a
Qa
*h I =
hI =
Wi
0
II
Qr
h II = h I
I
1-
a
Extract heat Q c from
cold reservoir at Tc
(below ambient)
a
Qa
1-
F TI
*h = h G1 - J
H TK
I
J
T K
T0
T0
Ta
T0
(heat pump)
Tr
(solar water heater)
A = Wi
A = Qr 1 -
F
GH
Amin = Q c
F T - 1I
GH T JK
0
Amin = Q c
c
*h I =
I
J
T K
T0
r
F T - 1I
GH T JK
0
c
Qc
*h I =
Wi
Qc
Qr
F T - 1I
*h = h G
H T JK
F T - 1I
G T JJ
*h = h G
GG 1 - T JJ
H TK
(Refrigerator—electric
motor driven)
(Refrigerator—heat
operated)
0
c
0
II
I
II
c
I
0
r
In the case of a heat pump, the task is to add heat Qa to a reservoir to be maintained at temperature Ta and the input shaft work is W i.
Q
COP = a = hI , say
Wi
Ta
Qa
Qa
(COP)max =
=
=
Ta - T0 Wi
A min
\
\
F
GH
Amin = Qa 1 -
h II =
Amin
* Strictly speaking, it is COP.
A
I
T JK
T0
a
F
GH
Qa 1 =
Wi
T0
Ta
I
JK
237
Available Energy, Availability and Irreversibility
F
GH
h II = h I 1 -
I
T JK
T0
(8.89)
a
Similarly, expressions of h I and h II can be obtained for other thermal tasks.
8.12.1
Matching End Use to Source
Combustion of a fuel releases the necessary energy for the tasks such as space
heating, process steam generation and heating in industrial furnaces. When the
products of combustion are at a temperature much greater than that required by a
given task, the end use is not well matched to the source and results in inefficient
utilization of the fuel burned. To illustrate this, let us consider a closed system
receiving a heat transfer Q r at a source temperature Tr and delivering Q a at a use
temperature Ta (Fig. 8.19). Energy is lost to the surroundings by heat transfer at a
rate Q l across a portion of the surface at Tl. At a steady state the energy and
availability rate balances become
∑
∑
∑
∑
∑
∑
Qr = Q a + Q l
F
GH
Qr 1 -
(8.90)
I = Q F1 - T I + Q F1 - T I + I
GH T JK GH T JK
T JK
T0
∑
0
a
r
0
∑
(8.91)
l
a
l
Ql
TI
Qr
Qa
Tr
Ta
Surroundings T0
System Boundary
Fig. 8.19 Efficient Energy Utilization from Second Law Viewpoint
∑
Equation (8.90) indicates that the energy carried in by heat transfer Q r is
∑
∑
either used, Q a , or lost to the surroundings, Q l . Then
∑
hI =
Qa
(8.92)
Qr
The value of hI can be increased by increasing insulation to reduce the loss. The
∑
∑
limiting value, when Q l = 0, is h I = 1 (100%).
Equation (8.91) shows that the availability carried into the system accompanying
the heat transfer Q r is either transferred from the system accompanying the heat
∑
∑
∑
∑
transfer Q a and Q l or destroyed by irreversibilities within the system, I .
238
Engineering Thermodynamics
Therefore,
F T I 1- T
GH T JK
T
h =
=h
F T I 1- T
Q G1 - J
T
H TK
∑
Qa 1 -
0
0
a
a
II
I
∑
0
r
(8.93)
0
r
r
Both hI and hII indicate how effectively the input is converted into the product.
The parameter hI does this on energy basis, whereas hII does it on an availability
basis.
For proper utilization of availability, it is desirable to make hI as close to unity as
practical and also a good match between the source and use temperatures, Tr and Ta.
Figure 8.20 demonstrates the second law efficiency against the use temperature Ta
for an assumed source temperature Tr = 2200 K. It shows that hII tends to unity
(100%) as Ta approaches Tr. The lower the Ta, the lower becomes the value of hII .
Efficiencies for three applications, viz., space heating at Ta = 320 K, process steam
generation at Ta = 480 K, and heating in industrial furnaces at Ta = 700 K, are indicated on the figure. It suggests that fuel is used far more effectively in the high
temperature use. An excessive temperature gap between Tr and Ta causes a low hI I
and an inefficient energy utilization. A fuel or any energy source is consumed
efficiently when the first user temperature approaches the fuel temperature. This
means that the fuel should first be used for high temperature applications. The heat
rejected from these applications can then be cascaded to applications at lower
temperatures, eventually to the task of, say, keeping a building warm. This is called
energy cascading and ensures more efficient energy utilization.
1.0
h II
h II
As Ta
0.662
1
Tr
Heating in Furnace
0.5
0.434
Process Steam Generation
System Boundary
0.072
Space Heating
0
0
300
500
1000
1500
2000
Ta (in K)
Fig. 8.20
8.12.2
Effect of Use Temperature Ta on The Second Law Efficiency
(Tr = 2200 K, T0 = 300 K, nl = 100%)
Further Illustrations of Second Law Efficiencies
Second law efficiency of different components can be expressed in different forms.
It is derived by using the exergy balance rate given as follows:
239
Available Energy, Availability and Irreversibility
(a) Turbines The steady state exergy balance (Fig. 8.21) gives:
∑
∑
∑
LM1 - T OP + a = W + a + I
T PQ
m
m
m MN
Q
0
f1
∑
s
f2
∑
∑
1
w
T
Ts
Q
2
Exergy Balance of a Turbine
Fig. 8.21
If there is no heat loss,
af – af =
1
2
∑
∑
W
I
+
∑
m
∑
(8.94)
∑
m
∑
W/ m
h II =
a f1 - a f 2
The second law efficiency,
(8.95)
(b) Compressor and Pump Similarly, for a compressor or a pump,
∑
∑
W
∑
m
= af – af +
hII =
and
2
1
I
∑
m
af2 - af2
∑
(8.96)
∑
- W /m
(c) Heat Exchanger Writing the exergy balance for the heat exchanger,
(Fig. 8.22)
LM
MN
 1-
OP Q - W
T PQ
T0
∑
∑
j
j
∑
∑
∑
∑
∑
C.V . + [ m h a f 1 + mh a f 3 ] - [ m h a f 2 + mc a f 4 ] - I C.V. = 0
If there is no heat transfer and work transfer,
∑
∑
∑
mh [af – af ] = m c [af – a f ] + I
1
2
4
3
(8.97)
∑
hII =
mc [ a f4 - a f3 ]
∑
m h [ a f1 - a f2 ]
(8.98)
240
Engineering Thermodynamics
Fig. 8.22
Exergy Balance of a Heat Exchanger
(d) Mixing of Two Fluids Exergy balance for the mixer (Fig. 8.23) gives:
LM1 - T OP Q + m a + m a = W
N TQ
∑
0
∑
∑
∑
1
s
2
f1
f2
∑
C.V. + m 3 a f3 + IC.V.
∑
∑
∑
∑
If the mixing is adiabatic and since WC.V. = 0 and m1 + m 2 = m 3 .
∑
∑
∑
m1 [af – af ] = m 2 [af – a f ] + I
1
3
3
(8.99)
2
and
∑
hII =
m 2 [ a f3 - a f2 ]
(8.100)
∑
m1 [a f1 - a f3 ]
Hot stream
m1
m3
3
m2
Cold stream
Fig. 8.23 Exergy Loss Due to Mixing
8.13
COMMENTS ON EXERGY
The energy of the universe, like its mass, is constant. Yet at times, we are bombarded
with speeches and articles on how to “conserve” energy. As engineers, we know
that energy is always conserved. What is not conserved is the exergy, i.e. the useful
work potential of the energy. Once the exergy is wasted, it can never be recovered.
When we use energy (electricity) to heat our homes, we are not destroying any
energy, we are merely converting it to a less useful form, a form of less exergy value.
The maximum useful work potential of a system at the specified state is called
exergy which is a composite property depending on the state of the system and the
surroundings. A system which is in equilibrium with its surroundings is said to be at
the dead state having zero exergy.
Available Energy, Availability and Irreversibility
241
The mechanical forms of energy such as KE and PE are entirely available energy
or exergy. The exergy (W) of thermal energy (Q) of reservoirs (TER) is equivalent to
the work output of a Carnot heat engine operating between the reservoir at tempera-
LM
N
ture T and environment at T0, i.e. W = Q 1 -
OP .
TQ
T0
The actual work W during a process can be determined from the first law. If the
volume of the system changes during a process, part of this work (W surr) is used to
push the surrounding medium at constant pressure p0 and it cannot be used for any
useful purpose. The difference between the actual work and the surrounding work
is called useful work, W u
Wu = W – Wsurr = W – p0(V2 – V1)
Wsurr is zero for cyclic devices, for steady flow devices, and for system with fixed
boundaries (rigid walls).
The maximum amount of useful work that can be obtained from a system as it
undergoes a process between two specified states is called reversible work, W rev. If
the final state of the system is the dead state, the reversible work and the exergy
become identical.
The difference between the reversible work and useful work for a process is
called irreversibility.
I = Wrev – W u = T0 Sgen
∑
∑
I = T0 Sgen
For a total reversible process, Wrev = W u and I = 0.
The first law efficiency alone is not a realistic measure of performance for engineering devices. Consider two heat engines, having e.g. a thermal efficiency of, say,
30%. One of the engines (A) is supplied with heat Q from a source at
600 K and the other engine (B) is supplied with the same amount of heat Q from a
source at 1000 K. Both the engines reject heat to the surroundings at 300 K.
300
(WA)rev = Q 1 = 0.5 Q, while (W A)act = 0.3Q
600
Similarly,
300
(WB)rev = Q 1 = 0.7 Q, and (WB)act = 0.3Q
1000
At first glance, both engines seem to convert the same fraction of heat, that they
receive, to work, thus performing equally well from the viewpoint of the first law.
However, in the light of second law, the engine B has a greater work potential (0.7 Q)
available to it and thus should do a lot better than engine A. Therefore, it can be said
that engine B is performing poorly relative to engine A, even though both have the
same thermal efficiency.
To overcome the deficiency of the first law efficiency, a second law efficiency hII
can be defined as the ratio of actual thermal efficiency to the maximum possible
thermal efficiency under the same conditions:
FG
H
FG
H
IJ
K
IJ
K
242
Engineering Thermodynamics
hI
hII = h
rev
So, for engine A, hII = 0.3/0.5 = 0.60
and for engine B, hII = 0.3/0.7 = 0.43
Therefore, the engine A is converting 60% of the available work potential (exergy)
to useful work. This is only 43% for the engine B. Therefore,
Wu
h act
hII =
=
(for heat engines and other work producing devices)
h rev
Wrev
hII =
Wrev
COP
=
(for refrigerators, heat pumps and other work absorbWu
COPrev
ing devices).
The exergies of a closed system (f) and a flowing fluid system (y) are given on
unit mass basis:
f = (u – u0) – T0 (s – s0) + p0 (v – v0) kJ/kg
y = (h – h 0) – T0(s – s0) +
V2
+ gz kJ/ kg
2
Reversible Work Expressions
(a) Cyclic Devices
W rev = h rev Q1 (Heat engines)
Q2
– Wrev =
(Refrigerators)
(COPrev ) Ref.
– Wrev =
Q2
(COPrev ) HP
(Heat pumps)
(b) Closed System
Wrev = U1 – U2 – T0(S 1 – S 2) + p0(V1 – V2)
= m (f 1 – f 2 )
(c) Steady Flow System (single stream)
Wrev = m
LMF h + V + gz - T s I - F h + V + gz - T s I OP
JK GH 2
JK PQ
MNGH 2
2
1
1
2
2
1
0 1
2
2
0 2
= m (y1 – y2 )
When the system exchanges heat with another reservoir at temperature T k other
than the atmosphere,
∑
∑
∑
F
GH
W rev = m (y1 – y 2 ) + Q k 1 -
I
T JK
T0
k
Available Energy, Availability and Irreversibility
243
The first law efficiency is defined as the ratio of energy output and energy input,
while their difference is the energy loss. Likewise, the second law efficiency is
defined as the ratio of exergy output and exergy input and their difference is irreversi
bility. By reducing energy loss, first law efficiency can be improved. Similarly, by
reducing irreversibilities, the second law efficiency can be enhanced.
Solved Examples
Example 8.1 In a certain process, a vapour, while condensing at 420°C,
transfers heat to water evaporating at 250°C. The resulting steam is used in a
power cycle which rejects heat at 35°C. What is the fraction of the available
energy in the heat transferred from the process vapour at 420°C that is lost due to
the irreversible heat transfer at 250°C?
Solution ABCD (Fig. 8.24) would have been the power cycle, if there was no
temperature difference between the vapour condensing and the water evaporating
and the area under CD would have been the unavailable energy. EFGD is the power
cycle when the vapour condenses at 420°C and the water evaporates at 250°C. The
unavailable energy becomes the area under DG. Therefore, the increase in unavailable energy due to irreversible heat transfer is represented by the area under CG.
Q1
T1 = 420 + 273 = 693 K
B
T
A
Q1
E
F
C
D
Ds
G
T¢1 = 250 + 273 = 523 K
T0 = 35 + 273 = 308 K
Increase in unavailable
energy
Ds¢
s
Fig. 8.24
Now
Q1 = T1 D S = T1¢ D S¢
D S¢
DS
=
T1
T 1¢
W = work done in cycle ABCD
= (T1 – T0) D S
W ¢ = Work done in cycle EFGD
= (T ¢1 – T0) DS ¢
244
Engineering Thermodynamics
The fraction of energy that becomes unavailable due to irreversible heat transfer
=
=
W - W¢
W
=
T0 ( D S ¢ - D S )
( T1 - T0 ) D S
T0 ( T1 - T 1¢ )
T 1¢ (T1 - T0 )
T0
=
FG DS¢ - 1IJ
H DS K
(T1 - T0 )
308 (693 - 523)
=
523 (693 - 308)
= 0.26
Example 8.2 In a steam boiler, hot gases from a fire transfer heat to water which
vaporizes at constant temperature. In a certain case, the gases are cooled from
1100 °C to 550°C while the water evaporates at 220°C. The specific heat of gases
is 1.005 kJ/kg-K, and the latent heat of water at 220°C, is 1858.5 kJ/kg. All the
heat transferred from the gases goes to the water. How much does the total entropy
of the combined system of gas and water increase as a result of the irreversible
heat transfer? Obtain the result on the basis of 1 kg of water evaporated.
If the temperature of the surroundings is 30°C, find the increase in unavailable
energy due to irreversible heat transfer.
∑
Solution Gas ( m g ) is cooled from state 1 to state 2 (Fig. 8.25). For reversible heat
transfer, the working fluid (w.f.) in the heat engine having the same cp would have
been heated along 2 – 1, so that at any instant, the temperature difference between
gas and the working fluid is zero. Then 1 – b would have been the expansion of the
working fluid down to the lowest possible temperature T0, and the amount of heat
rejection would have been given by the area abcd.
1
Q1
T
550∞C
2
mg
1100∞C
.
w.f
Increase in
unavailable
energy
Q1
220∞C
mw
Q2
mw
b
a
e
Ds gas
d
Ds H2O
c
T0 = 30 + 273
= 303 K
f
s
Fig. 8.25
When water evaporates at 220°C as the gas gets cooled from 1100°C to 550°C, the
resulting power cycle has an unavailable energy represented by the area aefd. The
increase in unavailable energy due to irreversible heat transfer is thus given by area
befc.
Available Energy, Availability and Irreversibility
245
Entropy increase of 1 kg water
Latent heat absorbed
(D S )water =
T
=
1858.5
(273 + 220)
= 3.77 kJ/kg-K
Q1 = Heat transferred from the gas
= Heat absorbed by water during evaporation
= mg cp (1100 – 550)
g
= 1 ¥ 1858.5 kJ
\
∑
m g cp =
1858.5
g
D Sgas =
550
z
Tg2
T g1
= 3.38 kJ/°C
d- Q
=
T
∑
= m g cp ln
g
Tg
Tg
z
T 82
T 81
dT
∑
m g cp
= 3.38 ln
g
T
823
1373
= – 3.38 ¥ 0.51
= – 1.725 kJ/K
\
D S total = (D S) water + (D S)gas
= 3.77 – 1.725 = 2.045 kJ/K
Increase in unavailable energy
= T0 (D S)total = 303 ¥ 2.045 = 620 kJ
Example 8.3 Calculate the available energy in 40 kg of water at 75°C with
respect to the surroundings at 5°C, the pressure of water being 1 atm.
Solution If the water is cooled at a constant pressure of 1 atm from 75°C to 5°C
(Fig. 8.26) the heat given up may be used as a source for a series of Carnot engines
each using the surroundings as a sink. It is assumed that the amount of energy
received by any engine is small relative to that in the source and the temperature of
the source does not change while heat is being exchanged with the engine.
T
273 + 75 = 348 K
T
p=
tm
1a
AE
ds
s
Fig. 8.26
U.E.
T0 = 278 K
246
Engineering Thermodynamics
Let us consider that the source has fallen to temperature T, at which level there
operates a Carnot engine which takes in heat at this temperature and rejects heat at
T0 = 278 K. If d s is the entropy change of water, the work obtainable is
dW = – m (T – T0) d s
where d s is negative.
cpd T
dW = – 40 (T – T0)
FG
H
= – 40 cp 1 -
\
T
IJ d T
TK
T0
With a very great number of engines in the series, the total work (maximum)
obtainable when the water is cooled from 348 K to 278 K would be
FG
H
278
W (max) = A.E. = – lim S 40 cp 1 348
=
z
348
278
FG
H
40 cp 1 -
T0
T
IJ d T
TK
T0
IJ dT
K
LM
N
= 40 cp (348 - 278) - 278 ln
OP
278 Q
348
= 40 ¥ 4.2 (70 – 62) = 1340 kJ
Q1 = 40 ¥ 4.2 (348 – 278) = 11,760 kJ
U.E. = Q1 – W (max)
= 11,760 – 1340 = 10,420 kJ
Example 8.4 Calculate the decrease in available energy when 25 kg of water
at 95°C mix with 35 kg of water at 35°C, the pressure being taken as constant and
the temperature of the surroundings being 15°C (cp of water = 4.2 kJ/kg K).
Solution The available energy of a system of mass m, specific heat cp, and at
temperature T, is given by
z FGH
T
A.E. = mcp
\
T0
1-
T0
T
IJ dT
K
(A.E.)25 = Available energy of 25 kg of water at 95°C
= 25 ¥ 4.2
LM
N
z
FG1 - 288IJ d T
H TK
273 + 95
273 + 15
= 105 (368 - 288) - 288 ln
368
288
OP = 987.49 kJ
Q
(A.E.)35 = Available energy of 35 kg of water at 35°C
Available Energy, Availability and Irreversibility
LM
N
= 147 (308 - 288) - 288 ln
247
OP = 97.59 kJ
288 Q
308
Total available energy
(A.E.)total = (A.E.)25 + (A.E.)35
= 987.49 + 97.59
= 1085.08 kJ
After mixing, if t is the final temperature
25 ¥ 4.2 (95 – t) = 35 ¥ 4.2 (t – 35)
\
t=
25 ¥ 95 + 35 ¥ 35
25 + 35
Total mass after mixing = 25 + 35 = 60 kg
= 60°C
(A.E.)60 = Available energy of 60 kg of water at 60°C
LM
N
= 4.2 ¥ 60 (333 - 288) - 288 ln
333
288
OP = 803.27 kJ
Q
\ Decrease in available energy due to mixing
= Total available energy before mixing
– Total available energy after mixing
= 1085.08 – 803.27 = 281.81 kJ
Example 8.5 The moment of inertia of a flywheel is 0.54 kg ◊ m 2 and it rotates at
a speed 3000 RPM in a large heat insulated system, the temperature of which is
15°C. If the kinetic energy of the flywheel is dissipated as frictional heat at the
shaft bearings which have a water equivalent of 2 kg, find the rise in the temperature of the bearings when the flywheel has come to rest. Calculate the greatest
possible amount of this heat which may be returned to the flywheel as high-grade
energy, showing how much of the original kinetic energy is now unavailable.
What would be the final RPM of the flywheel, if it is set in motion with this available energy?
Solution
Initial angular velocity of the flywheel
w1 =
2p N1
=
2p ¥ 3000
60
60
Initial available energy of the flywheel
= (K.E.)initial =
1
2
= 314.2 rad /s
Iw 12
= 0.54 kg m2 ¥ (314.2)2
rad2
s2
= 2.66 ¥ 10 Nm = 26.6 kJ
4
248
Engineering Thermodynamics
When this K.E. is dissipated as frictional heat, if D t is the temperature rise of the
bearings, we have
water equivalent of the bearings ¥ rise in temperature = 26.6 kJ
\
Dt =
26.6
2 ¥ 4187
.
= 3.19°C
\ Final temperature of the bearings
tf = 15 + 3.19 = 18.19°C
The maximum amount of energy which may be returned to the flywheel as highgrade energy is
A.E. = 2 ¥ 4.187
z FGH
29119
.
288
1-
IJ
K
288
dT
T
LM
N
= 2 ¥ 4.187 (29119
. - 288) - 288 ln
OP
288 Q
29119
.
= 0.1459 kJ
The amount of energy rendered unavailable is
U.E. = (A.E.)initial – (A.E.) returnable as high grade energy
= 26.6 – 0.1459
= 26.4541 kJ
Since the amount of energy returnable to the flywheel is 0.146 kJ, if w 2 is the final
angular velocity, and the flywheel is set in motion with this energy
0.146 ¥ 103 =
1
2
¥ 0.54 w 22
146
\
w 22 =
\
w 2 = 23.246 rad/s
0.27
= 540.8
If N 2 is the final RPM of the flywheel
w 2 = 23.246 =
or
N2 =
2p N2
60
23.246 ¥ 60
2¥p
= 222 RPM
Example 8.6 Two kg of air at 500 kPa, 80°C expands adiabatically in a closed
system until its volume is doubled and its temperature becomes equal to that of the
surroundings which is at 100 kPa, 5°C. For this process, determine (a) the maximum work, (b) the change in availability, and (c) the irreversibility. For air, take
cv = 0.718 kJ/kg K, u = cvT where cv is contant, and pV = mRT where p is pressure
in kPa, V volume in m3, m mass in kg, R a constant equal to 0.287 kJ/kg K, and T
temperature in K.
249
Available Energy, Availability and Irreversibility
From the property relation
TdS = dU + pdV
the entropy change of air between the initial and final states is
2 mc dT
2 m RdV
2
v
+
dS =
1
1
1
T
V
T2
V
or
S2 – S 1 = mcv ln
+ mR ln 2
T1
V1
Solution
z z
z
From Eq. (8.32),
Wmax = (U1 – U2) – T0 (S1 – S 2)
LM
MN
F
GH
= m cv ( T1 - T2 + T0 cv ln
LM
N
= 2 0.718 (80 - 5) + 278
T2
T1
+ R ln
V2
V1
I OP
JK PQ
F 0.718 ln 278 + 0.287 ln 2 I OP
H
353
1K Q
= 2 [53.85 + 278 (– 0.172 + 0.199)]
= 2 (53.85 + 7.51) = 122.72 kJ
From Eq. (8.42), the change in availability
= f1 – f 2
= (U1 – U2) – T0(S1 – S2) + p0(V1 – V2)
= Wmax + p0(V1 – V2)
= 122.72 + p0(V1 – 2V1)
= 122.72 – 100 ¥
2 ¥ 0.287 ¥ 353
500
= 82.2 kJ
The irreversibility
I = Wmax, useful – Wact
From the first law,
\
Wact = Q – DU = – DU = U1 – U2
I = U1 – U2 – T0(S 1 – S 2) – U1 + U2
= T0(S2 – S1)
= T0(DS)system
For adiabatic process, (DS) surr = 0
LM
MN
I = T0 mcv ln
LM
N
T2
T1
+ m R ln
= 278 ¥ 2 0.718 ln
OP
V PQ
V2
1
278
+ 0.287 ln 2
OP
Q
353
= 278 ¥ 2 (– 0.172 + 0.199) = 15.2 kJ
250
Engineering Thermodynamics
Example 8.7 Air expands through a turbine from 500 kPa, 520°C to 100 kPa,
300°C. During expansion 10 kJ/kg of heat is lost to the surroundings which is at
98 kPa, 20°C. Neglecting the K.E. and P.E. changes, determine per kg of air (a) the
decrease in availability, (b) the maximum work and (c) the irreversibility. For air,
take cp = 1.005 kJ/kg K, h = cp T where cp is constant, and the p, V and T relation
as in Example 8.6.
From the property relation
TdS = dH – Vdp
the entropy change of air in the expansion process is
Solution
2
2
mc p dT
1
T
z z
dS =
1
or
S2 – S 1 = mcp ln
T2
T1
-
z
2 m Rd p
p
1
– mR ln
p2
p1
For 1 kg of air,
s2 – s 1 = cp ln
T2
T1
– R ln
p2
p1
From Eq. (8.30), the change in availability
y1 – y2 = b1 – b2
= (h1 – T0s1) – (h 2 – T0s 2)
= (h1 – h2) – T0(s1 – s2)
F
GH
= cp (T1 – T2) – T0 R ln
p2
p1
- cp ln
FG
H
= 1.005 (520 – 300) – 293 0.287 ln
I
T JK
T2
1
1
5
= 1.005 ¥ 220 – 293 (0.3267 – 0.4619)
= 221.1 + 39.6 = 260.7 kJ/kg
The maximum work is
Wmax = change in availability = y1 – y2
= 260.7 kJ/kg
From S.F.E.E.,
Q + h1 = W + h2
W = (h1 – h 2) + Q
= cp (T1 – T2) + Q
= 1.005 (520 – 300) – 10 = 211.1 kJ/kg
The irreversibility
I = Wmax – W
= 260.7 – 211.1 = 49.6 kJ/kg
- 1.005 ln
IJ
793K
573
Available Energy, Availability and Irreversibility
251
Alternatively,
I = T0 (DSsystem + D Ssurr)
FG
H
= 293 1.005 ln
573
- 0.287 ln
IJ
5 293K
1
+
10
793
= 293 ¥ 0.1352 + 10 = 49.6 kJ/kg
Example 8.8 An air preheater is used to cool the products of combustion from a
furnace while heating the air to be used for combustion. The rate of flow of products is 12.5 kg/s and the products are cooled from 300 to 200°C, and for the
products at this temperature cp = 1.09 kJ/kg K. The rate of air flow is 11.5 kg/s, the
initial air temperature is 40°C, and for the air cp = 1.005 kJ/kg K. (a) Estimate the
initial and final availability of the products. (b) What is the irreversibility for the
process? (c) If the heat transfer from the products occurs reversibly through heat
engines, what is the final temperature of the air? What is the power developed by
the heat engine? Take T0 = 300 K and neglect pressure drop for both the fluids and
heat transfer to the surroundings.
Solution
(a)
y1 = initial availability of the products
= (h1 – h0) – T0 (s1 – s 0)
= cp (Tg1 – T0) – T0c p ln
g
g
Tg1
T0
= 1.09 (573 – 300) – 300 ¥ 1.09 ln
573
300
= 297.57 – 211.6 = 85.97 kJ/kg
y2 = final availability of the products
= (h2 – h0) – T0(s2 – s0 )
= 1.09 (473 – 300) – 300 ¥ 1.09 ¥ ln
473
300
= 188.57 – 148.89 = 39.68 kJ/kg
(b) Decrease in availability of the products
= y1 – y2
= (h1 – h2) – T0(s1 – s2)
= 1.09 (573 – 473) – 300 ¥ 1.09 ln
573
473
= 109 – 62.72 = 46.28 kJ/kg
By making an energy balance for the air preheater [Fig. 8.27 (a)].
∑
∑
m g cp (Tg1 – Tg2) = m a cpa (Ta – Ta )
g
2
1
12.5 ¥ 1.09 (573 – 473) = 11.15 ¥ 1.005 (Ta – 313)
2
252
Engineering Thermodynamics
\
Ta =
2
12.5 ¥ 109
115
. ¥ 1005
.
+ 313 = 430.89 K
Products
of Combustion
mg
F
u
r
n
a
c
e
Fuel
Tg1
Air
Preheater
Tg2 Exhaust (mg)
ma
Preheated
Air
Ta2
Ta1 Air (ma)
Insulation
Fig. 8.27 (a)
Increase in availability for air
= y2 – y1
= (h2 – h1) – T0(s2 – s1)
= cpa (Ta2 – Ta1) – T0cpa ln
Ta2
Ta1
= 1.005 ¥ (430.89 – 313) – 300 ¥ 1.005 ln
430.89
313
= 118.48 – 96.37 = 22.11 kJ/kg
\ Irreversibility of the process
= 12.5 ¥ 46.28 – 11.5 ¥ 22.11
= 578.50 – 254.27 = 324.23 kW
(c) Let us assume that heat transfer from the products to air occurred through
heat engines reversibly as shown in Fig. 8.27 (b).
Tg1
Tg2
Tg1
mg
mg
Ta2
Q1
T
Q1
W
E
mg
Q1
Q2
W
Q1
E
ma
Q2
Ta2
ma
Q2
Q2
Ta1
Ta1
ma
Fig. 8.27
Tg2
W
A0 or L
(b)
Available Energy, Availability and Irreversibility
253
For reversible heat transfer,
∑
D Suniv = 0
∑
∑
D Ssys + D Ssurr = 0
∑
∑
D Sgas + D Sair = 0
∑
∑
D S gas = – D S air
∑
mg c pg ln
12.5 ¥ 1.09 ln
\
Tg 2
T g1
473
∑
= ma c pa ln
Ta2
Ta1
= – 11.5 ¥ 1.005 ln
573
Ta = 392.41 K
Ta2
313
2
Rate of heat supply from the gas to the working fluid in the heat engine,
∑
∑
Q1 = mg cp (Tg – Tg )
g
1
2
= 12.5 ¥ 1.09 (573 – 473) = 1362.50 kW
Rate of heat rejection from the working fluid in the heat engine to the air,
∑
∑
Q2 = ma cp (Ta – Ta )
a
2
1
= 11.5 ¥ 1.005 (392.41 – 313) = 917.78 kW
Total power developed by the heat engine
∑
∑
∑
W = Q1 – Q2 = 1362.50 – 917.78 = 444.72 kW
Example 8.9 A gas is flowing through a pipe at the rate of 2 kg/s. Because of
inadequate insulation the gas temperature decreases from 800 to 790°C between
two sections in the pipe. Neglecting pressure losses, calculate the irreversibility
rate (or rate of energy degradation) due to this heat loss. Take T0 = 300 K and a
constant cp = 1.1 kJ/kg K.
For the same temperature drop of 10°C when the gas cools from 80°C to 70°C
due to heat loss, what is the rate of energy degradation? Take the same values of T0
and cp. What is the inference you can draw from this example?
Solution
∑
Q
Sgen = Ssys T0
∑
∑
∑
∑
= m sys (s2 – s 1) –
mc p ( T2 - T1 )
T0
Irreversibility rate = rate of energy degradation
= rate of exergy loss
254
Engineering Thermodynamics
.
.
I = T0 Sgen
.
.
= m T0 (s2 – s1) – m c p (T2 – T1)
LM
N
T
.
= mcp (T1 - T2 ) - T0 ln 1
T2
OP
Q
LM
N
= 2 ¥ 1.1 (1073 - 1063) - 300 ln
OP
Q
1073
= 15.818 kW
1063
When the same heat loss occurs at lower temperature
.
353
I = 2 ¥ 1.1 (353 - 343) - 300 ln
= 3.036 kW
343
LM
N
OP
Q
It is thus seen that irreversibility rate or exergy destruction is more when the
same heat loss occurs at higher temperature. Irreversibility rate decreases as the
temperature of the gas decreases. Quantitatively, the heat loss may be the same, but
qualitatively, it is different.
Example 8.10 An ideal gas is flowing through an insulated pipe at the rate of 3
kg/s. There is a 10% pressure drop from inlet to exit of the pipe. What is the rate of
exergy loss because of the pressure drop due to friction? Take R = 0.287 kJ/kg K
and T0 = 300 K.
Solution
Rate of entropy generation from Eq. (8.68),
.
. Dp
Sgen = m R
p1
= 3 ¥ 0.287
0.10 p 1
= 0.0861 kW/K
p1
Rate of exergy loss
∑
∑
I = T0 Sgen
= 300 ¥ 0.0861 = 25.83 kW
Example 8.11 Water at 90°C flowing at the rate of 2 kg/s mixes adiabatically
with another stream of water at 30°C flowing at the rate of 1 kg/s. Estimate the
entropy generation rate and the rate of exergy loss due to mixing. Take T0 = 300 K.
Solution
∑
∑
∑
m = m1 + m 2 = 2 + 1 = 3 kg/s
Here
∑
x=
t=
m1
∑
m
T2
T1
=
=
2
3
= 0.67
303
363
= 0.835
Available Energy, Availability and Irreversibility
255
From Eq. (8.76),
.
x + t (1 - x)
Sgen = mc p ln
t 1- x
0.67 ¥ 0.835 ¥ 0.33
= 3 ¥ 4.187 ln
0.835 0.33
0.94555
= 12.561 ln
= 0.0442 kW/K
0.94223
∑
Rate of exergy loss due to mixing
∑
∑
I = T0 Sgen
= 300 ¥ 0.0442 = 13.26 kW
Alternatively,
Equilibrium temperature after mixing,
t=
=
m1t 1 + m 2t 2
m1 + m 2
2 ¥ 90 + 1 ¥ 30
2 +1
= 70°C
.
343
343
D Suniv = Sgen = 2 ¥ 4.187 ln
+ 1 ¥ 4.187 ln
363
303
∑
= 0.0447 kW/K
\
∑
I = 300 ¥ 0.0447 = 13.41 kW
Example 8.12 By burning a fuel the rate of heat release is 500 kW at 2000 K.
What would be the first law and the second law efficiencies if (a) energy is absorbed in a metallurgical furnace at the rate of 480 kW at 1000 K, (b) energy is
absorbed at the rate of 450 kW for generation of steam at 500 K, and (c) energy is
absorbed in a chemical process at the rate of 300 kW at 320 K? Take T0 = 300 K.
(d) Had the energy absorption rate been equal to 450 kW in all these three cases,
what would have been the second law efficiencies? What is the inference that you
can draw from this example?
.
.
Solution If Q r is the rate of heat release at temperature Tr and Q a the rate of heat
absorption at temperature Ta, then
T0
Q
Ta
hI = a and hI I = h I
T
Qr
1- 0
Tr
∑
∑
1-
256
Engineering Thermodynamics
(a) Metallurgical furnace
hI =
480
500
¥ 100 = 96%
300
1000
h II = 0.96
¥ 100 = 79%
300
12000
1-
(b) Steam generation
hI =
450
500
¥ 100 = 90%
300
500 ¥ 100 = 42.3%
h II = 0.90
300
12000
1-
(c) Chemical process
hI =
300
500
¥ 100 = 60%
300
320 ¥ 100 = 4.41%
h II = 0.60
300
12000
(d) In all the three cases, hI would remain the same, where
1-
hI =
450
500
¥ 100 = 0.90
300
1000
h I I (a) = 0.90 ¥
¥ 100 = 74.11%
300
12000
1-
300
500 ¥ 100 = 42.3%
h I I (b) = 0.90 ¥
300
12000
1-
300
320 ¥ 100 = 6.61%
h II (c) = 0.90 ¥
300
12000
1-
∑
∑
It is seen that as the energy loss ( Qr - Qa ) increases, the first law efficiency
decreases. For the same heat loss, however, as the temperature difference between
the source and the use temperature increases, the second law efficiency decreases,
or in other words, the rate of exergy loss increases.
257
Available Energy, Availability and Irreversibility
Example 8.13 A system undergoes a power cycle while receiving energy Q 1 at
temperature T1 and discharging energy Q 2 at temperature T2. There are no other
heat transfers.
(a) Show that the thermal efficiency of the
T1
cycle can be expressed as:
h = 1-
T2
-
T2 I
Q1
T1 T0 Q 1
where T0 is the ambient temperature and
I is the irreversibility of the cycle.
(b) Obtain an expression for the maximum
theoretical value for the thermal efficiency.
(c) Derive an expression for the irreversibility for which no network is developed by
the cycle. What conclusion do you derive
from it?
Solution
E
W = Q1 – Q2
Q2
T2
Fig. 8.28
An availability balance for the cycle gives (Fig. 8.28)
F
GH
(DA)cycle = 0 = 1 -
I Q – F1 - T I Q – W – I
GH T JK
T JK
T0
0
1
2
1
2
since each property is restored to its initial state.
Since
Q2 = Q1 – W,
F
GH
I Q – F1 - T I (Q – W ) – W – I
GH T JK
T JK
T LF
T I F
T IO
TI
W=
M
1 - J - G1 - J P Q G
T MH
N T K H T K PQ T
0 = 1-
T0
0
1
1
1
2
2
0
0
2
1
0
h=
1
W
Q1
=1–
T2
T1
2
-
0
T2 I
Proved.
T0 Q1
(b) When I = 0,
hmax = 1 –
T2
T1
(c) When W = 0
h =0=1–
I = T0
T2
T1
-
T2 I
T0 Q1
LM 1 - 1 OP Q = T LM Q - Q OP = T S
MN T T PQ
MN T T PQ
1
2
1
1
1
2
1
0
0
gen
The heat transfer Q1 from T1 to T2 takes place through a reversible engine, and
the entire work is dissipated in the brake, from which an equal amount of heat is
258
Engineering Thermodynamics
rejected to the reservoir at T2. Heat transfer through a finite temperature difference is
thus equivalent to the destruction of its exergy. (See Art. 8.10.1(a)).
Example 8.14 A compressor operating at steady state takes in 1 kg/s of air at 1
bar and 25°C and compresses it to 8 bar and 160°C. Heat transfer from the compressor to its surroundings occurs at a rate of 100 kW. (a) Determine the power
input in kW. (b) Evaluate the second law efficiency for the compressor. Neglect KE
and PE changes. Take T0 = 25°C and P0 = 1 bar.
Solution
SFEE for the compressor (Fig. 8.29) gives:
∑
∑
∑
W = Q + m (h1 – h2) = – 100 + 1 ¥ 1.005 (25 – 160) = – 235.7 kW
2
C
W
Q
1
Air
Fig. 8.29
Exergy balance for the compressor gives:
IJ - W - m a = I
TK
F T IJ + I
- W = m (a – a ) – Q G1 H TK
∑
∑
FG
H
m a f1 + Q 1 -
∑
T0
∑
f2
∑
∑
f2
∑
∑
∑
0
f1
∑
hII =
m (a f 2 - a f1 )
∑
W
a f – a f = h2 – h1 – T0(s2 – s1)
2
1
F
GH
= cp (T2 – T1) – T0 c p ln
T2
T1
- R ln
FG
H
= 1.005 (160 – 25) – 298 1.005 ln
= 200.95 kJ/kg
hII =
200.95
235.7
= 0.853 or, 85.3%
I
p JK
p2
1
433
298
IJ
K
- 0.287 ln 8
Available Energy, Availability and Irreversibility
Example 8.15
259
Determine the exergy of 1 m3 of complete vacuum.
Solution
f = U – U0 + p0(V – V0) – T0(S – S0 )
= H – H0 – V (p – p0) – T0(S – S0)
Since a vacuum has zero mass,
U = 0, H0 = 0, and S = 0
If the vacuum was reduced to the dead state
U0 = 0, H0 = S0 = 0 and V0 = 0
The pressure p for the vacuum is zero.
p0 = 1 bar = 100 kPa and V = 1 m3
But
f = p0V = 100
kN
¥ 1 m3 = 100 kJ
m2
If an air motor operates between the atmosphere and the vacuum, this is the
maximum useful work obtainable. Therefore, the vacuum has an exergy or work
potential.
Example 8.16 A mass of 1000 kg of fish initially at 1 bar, 300 K is to be cooled
to – 20°C. The freezing point of fish is – 2.2°C, and the specific heats of fish below
and above the freezing point are 1.7 and 3.2 kJ/kg K respectively. The latent heat
of fusion for the fish can be taken as 235 kJ/kg. Calculate the exergy produced in
the chilling process. Take T0 = 300 K and p0 = 1 bar.
Solution
Exergy produced= H2 – H1 – T0(S2 – S1)
With reference to Fig. 8.30
T1 = T0
1
300K
Cl
b
2
a
270.8K
253K
s
Fig. 8.30
H1 – H2 = 1000 [1.7 (270.8 – 253) + 235 + 3.2 (300 – 270.8)]
= 1000 [1.7 ¥ 17.8 + 235 + 3.2 ¥ 29.2]
= 1000 [30.26 + 235 + 93.44] = 358.7 MJ
260
Engineering Thermodynamics
H2 – H1 = – 358.7 MJ
LM
N
S1 – S2 = 1000 1.7 ln
270.8
+
235
+ 3.2 ln
300
OP
Q
253
270.8
270.8
= – 1000 [0.1156 + 0.8678 + 0.3277] = 1.311 MJ/K
S2 – S1 = – 1.311 MJ/K
Exergy produced = – 358.7 + 300 ¥ 1.311
= – 358.7 + 393.3 = 34.6 MJ or 9.54 kWh
Example 8.17 A quantity of air initially at 1 bar, 300 K undergoes two types of
interactions: (a) it is brought to a final temperature of 500 K adiabatically by
paddle-wheel work transfer, (b) the same temperature rise is brought about by
heat transfer from a thermal reservoir at 600 K. Take T0 = 300 K, p0 = 1 atm.
Determine the irreversibility (in kJ/kg) in each case and comment on the results.
Fig. 8.31
Solution
Case (a): As shown in the above figures (Fig. 8.31)
500
T
D suniv = sgen = cv ln 2 = 0.718 ln
= 0.367 kJ/kg K
300
T1
I = 300 ¥ 0.367 = 110.1 kJ/kg
Case (b):
Q = m cv (T2 – T1)
= 1 ¥ 0.718 (500 – 300) = 143.6 kJ/kg
143.6
Q2
= 0.367 –
= 0.1277 kJ/kg K
T
600
I = 300 ¥ 0.1277 = 38.31 kJ/kg
D suniv = s2 – s1 –
Comment:
The irreversibility in case (b) is less than in case (a).
Q
Ia = T0(s2 – s1), I b = T0 (s2 – s1) –
T
Q
Ia – Ib =
T
The irreversibility in case (b) is always less than in case (a) and the two
values would approach each other only at high reservoir temperature, i.e.
Ia Æ Ib as T Æ •
Example 8.18 Steam enters a turbine at 30 bar, 400°C (h = 3230 kJ/kg,
s = 6.9212 kJ/kg K) and with a velocity of 160 m/s. Steam leaves as saturated
261
Available Energy, Availability and Irreversibility
vapour at 100°C (h = 2676.1 kJ/kg, s = 7.3549 kJ/kg K) with a velocity of
100 m/s. At steady state the turbine develops work at a rate of 540 kJ/kg. Heat
transfer between the turbine and its surroundings occurs at an average outer
surface temperature of 500 K. Determine the irreversibility per unit mass. Give an
exergy balance and estimate the second law efficiency of the turbine. Take p0 = 1
atm, T0 = 298 K and neglect PE effect.
Solution By exergy balance of the control
volume (Fig. 8.32),
T0
af = W + Q 1 + af + I
1
2
TB
where af is the exergy transfer per unit
mass.
FG
H
W
T0
IJ
K
ST
C.V.
af1
TB
af 2
Q
Fig. 8.32
F TI
I = a – a – W – Q G1 - J
H TK
0
f1
f2
B
= (h1 – h2) – T0(s1 – s2) +
V12 - V22
2
FG
H
– W –Q 1-
= (3230.9 – 2676.1) – 298(6.9212 – 7.3549) +
FG
H
¥ 10–3 – 540 – Q 1 -
IJ
T K
T0
B
160 2 - 1002
2
IJ
500 K
298
= 151.84 – Q(0.404)
(1)
By SFEE,
h1 +
V12
2
= W + h2 + Q +
Q = (h1 – h2) +
V22
2
V12 - V22
2
= (3230.9 – 2676.1) +
–W
160 2 - 1002
2
¥ 10–3 – 540 = 22.6 kJ/kg.
From Eq. (1),
I = 151.84 – 22.6 ¥ 0.404 = 142.71 kJ/kg
Net exergy transferred to turbine
a f – a f = 691.84 kJ/kg
1
2
Work = 540 kJ/kg
Exergy destroyed = I = 142.71 kJ/kg
Exergy transferred out accompanying heat transfer
= 22.6 ¥ 0.404 = 9.13 kJ/kg
262
Engineering Thermodynamics
Exergy Balance
Exergy transferred
Exergy utilized
691.84 kJ/kg
Work = 540 kJ/kg (78%)
Destroyed = 142.71 kJ/kg (20.6%)
Transferred with heat = 9.13 kJ/kg (1.3%)
691.84 kJ/kg
540
Second law efficiency, h I I =
= 0.78 or 78%
69184
.
Example 8.19 A furnace is heated by an electrical resistor. At steady state,
electrical power is supplied to the resistor at a rate of 8.5 kW per metre length to
maintain it at 1500 K when the furnace walls are at 500 K. Let T0 = 300 K (a) For
the resistor as the system, determine the rate of availability transfer accompanying
heat and the irreversibility rate, (b) For the space between the resistor and the
walls as the system, evaluate the irreversibility rate.
Solution
Case (a): At steady state for the resistor (Fig. 8.33).
◊
◊
◊
◊
Q = DU + W = W = 8.5 kW
Resistor at 500 K
Space
I
I
Q
Furnace walls at 500 K
Fig. 8.33
Availability rate balance gives
dA
F
H
= 1-
dt
I
K
FG
H
IJ
K
dV
T0
–I=0
Q + W - p0
dt
T
∑
∑
I = Rate of irreversibility
F
H
= 1-
FG
H
IJ
K
300
T0
(– 8.5) + 8.5 = 1.7 kW
Q + W = 1T
1500
I
K
∑
∑
Rate of availability transfer with heat
F
H
= 1-
FG
H
IJ
K
300
T0
(– 8.5) = – 6.8 kW
Q = 1T
1500
I
K
∑
Available Energy, Availability and Irreversibility
Case (b): Steady state,
dA
263
F T I Q – F1 - T I Q - W - I = 0
H TK H TK
dT
F 300 IJ 8.5 – FG1 - 300 IJ 8.5 = 6.8 – 3.4 = 3.4 kW
I = G1 H 1500 K H 500 K
= 1-
∑
0
0
∑
∑
∑
◊
Example 8.20 Air enters a compressor at 1 bar, 30°C, which is also the state of
the environment. It leaves at 3.5 bar, 141°C and 90 m/s. Neglecting inlet velocity
and P.E. effect, determine (a) whether the compression is adiabatic or polytropic,
(b) if not adiabatic, the polytropic index, (c) the isothermal efficiency, (d) the
minimum work input and irreversibility, and (d) the second law efficiency. Take cp
of air = 1.0035 kJ/kg K.
Solution
(a) After isentropic compression
T2 s
T1
=
LM p OP
MN p PQ
( g - 1)/ g
2
1
T2s = 303 (3.5)0.286 = 433.6 K = 160.6°C
Since this temperature is higher than the given temperature of 141°C, there is heat
loss to the surroundings. The compression cannot be adiabatic. It must be polytropic.
(b)
Lp O
= M P
T
MN p PQ
T2
2
1
1
141 + 273
30 + 273
= 1.366 =
log 1.366 =
1–
1
n
( n - 1)/ n
=
n -1
n
0.135
0.544
LM 3.5 OP
N1Q
( n - 1)/ n
log 3.5
= 0.248
n = 1.32978 = 1.33
(c) Actual work of compression
V22
90
Wa = h1 – h2 –
= 1.0035 (30 – 141) –
¥ 10 –3 = – 115.7 kJ/kg
2
2
Isothermal work
2
V22
p 2 V22
WT = v d p –
= – RT1 ln
p1
2
2
1
z
= – 0.287 ¥ 303 ln (3.5) –
90 2
2
¥ 10–3 = – 113 kJ/kg
264
Engineering Thermodynamics
Isothermal efficiency:
hT =
WT
Wa
=
113
115.7
= 0.977 or 97.7%
(d) Decrease in availability or exergy:
y1 – y2 = h1 – h2 – T0(s1 – s2) +
V12 - V22
2
L p - c ln T OP - V
= c (T – T ) – T MR ln
T PQ 2
MN p
= 1.0035 (30 – 141)
414 O 90
L
– 303 M0.287 ln 3.5 - 1.0035 ln
303 PQ 2000
N
2
p
1
2
2
0
2
2
p
1
1
2
= – 101.8 kJ/kg
Minimum work input = – 101.8 kJ/kg
Irreversibility,
I = Wrev – Wa
= – 101.8 – (– 115.7) = 13.9 kJ/kg
(e) Second law efficiency,
h II =
Minimum work input
Actual work input
=
1018
.
115.7
= 0.88 or 88%
Summary
The maximum work obtainable Wmax from a certain heat input in a cyclic heat
Ê T ˆ
engine Q1 is called the available energy (A.E.), Á1 - 0 ˜ Q1 where T0 is the
Ë T1 ¯
temperature of the surroundings. The unavailable energy U.E. is the product
of the lowest temperature of heat rejection and the change of entropy of the
system during the process of supplying heat. The A.E. is also known as
exergy and the U.E. as anergy.
Whenever heat is transferred through a finite temperature difference there
is a decrease in the availability of energy so transferred. Energy is said to be
degraded each time it flows through a finite temperature difference. The
available energy or exergy of a fluid at temperature is given by
È
Í
Î
È
T
Wmax = mcp Í (T – T0 ) – T0 ln
T0
Î
where m is the mass of the fluid. The higher the temperature T, the more the
exergy. The exergy value provides a useful measure of the energy quality.
The first law states that energy is always conserved quantitywise, the
second law emphasizes that energy always degrades qualitywise.
Available Energy, Availability and Irreversibility
265
The work done by a closed system in a reversible process between two
states is always more than that done by any irreversible process between the
same states. The reversible work in a steady flow process is given by
Ê
ˆ Ê
ˆ
V1z
V2
+ gZ1˜ - Á h2 - T0 s2 + 2 + gZ2 ˜
Wmax = ÁË h1 - T0 s1 +
¯ Ë
¯
2
2
For a closed system it is
Wmax = (u1 – T0s1) – (u2 – T0 s2)
A certain portion of the work done by a closed system is spent in pushing the
atmosphere. The useful work becomes
(Wu)max = Wmax – p0 (V2 – V1)
When a system undergoes spontaneously due to a certain potential
difference, the process must terminate when the system reaches the dead
state, i.e., the pressure and temperature of the surroundings. The work done
by a system between two equilibrium states at the same temperature as that of
the surroundings is equal to or less than the change in Halmholtz function.
(WT)max £ (F1 – F2)T
The decrease in the Gibbs function of a system sets an upper limit to the
useful work done at the same pressure and temperature as those of the
surroundings
(Wu, max)p, T = (G1 – G2)p,T
The irreversibility of a process, I, is equal to
I = Wmax – W
The Guoy–Stodola theorem states that the rate of loss of exergy in a process
or irreversibility rate is given by
o
o
o
o
I = T0 D S = T0 (D S yst + D S surr).
While energy is always conserved, exergy is always destroyed by irreversibilities.
Energy in – Energy out = 0
Exergy in – Exergy out = Exergy destroyed
The second law efficiency is given by
hII =
where
W
Exergy output
=
Wmax
Exergy input
I = Wmax – W = T0 Sgen.
Review Questions
8.1 What do you understand by high grade energy and low grade energy?
8.2 What is available energy and unavailable energy?
8.3 Who propounded the concept of availability?
266
Engineering Thermodynamics
8.4 What is the available energy referred to a cycle?
8.5 Show that there is a decrease in available energy when heat is transferred
through a finite temperature difference.
8.6 Deduce the expression for available energy from a finite energy source at
temperature T when the environmental temperature is T0 .
8.7 What do you understand by exergy and anergy?
8.8 What is meant by quality of energy?
8.9 Why is exergy of a fluid at a higher temperature more than that at a lower
temperature?
8.10 How does the exergy value provide a useful measure of the quality of energy?
8.11 Why is the second law called the law of degradation of energy?
8.12 Energy is always conserved, but its quality is always degraded. Explain.
8.13 Why is the work done by a closed system in a reversible process by interacting only with the surroundings the maximum?
8.14 Show that equal work is done in all reversible processes between the same
end states of a system if it exchanges energy only with the surroundings.
8.15 Give the general expression for the maximum work of an open system which
exchanges heat only with the surroundings.
8.16 What do you understand by Keenan function?
8.17 Give the expression for reversible work in a steady flow process under a given
environment.
8.18 Give the expression for reversible work done by a closed system if it interacts
only with the surroundings.
8.19 What do you understand by ‘useful work’? Derive expressions for useful
work for a closed system and a steady flow system which interact only with
the surroundings.
8.20 What are the availability functions for a: (a) closed system, (b) steady flow
system?
8.21 What do you understand by the dead state?
8.22 What is meant by availability?
8.23 Give expressions for availabilities of a closed system and a steady flow open
system.
8.24 What are Helmholtz function and Gibbs function?
8.25 What is the availability in a chemical reaction if the temperature before and
after the reaction is the same and equal to the temperature of the surroundings?
8.26 When is the availability of a chemical reaction equal to the decrease in the
Gibbs function?
8.27 Derive the expression for irreversibility or exergy loss in a process executed
by: (a) a closed system, (b) a steady flow system, in a given environment.
8.28 State and explain the Gouy-Stodola theorem.
8.29 How is heat transfer through a finite temperature difference equivalent to the
destruction of its availability?
8.30 Considering the steady and adiabatic flow of an ideal gas through a pipe,
show that the rate of decrease in availability or lost work is proportional to the
pressure drop and the mass flow rate.
8.31 What do you understand by Grassman diagram?
Available Energy, Availability and Irreversibility
267
8.32 What is entropy generation number?
8.33 Define the second law efficiency. How is it different from the first law efficiency in the case of a simple power plant?
8.34 Derive the second law efficiency for: (a) a solar water heater, and (b) a heat
pump.
8.35 What is meant by energy cascading? How is it thermodynamically efficient?
8.36 Why is exergy always a positive value? Can it be negative?
8.37 Why and when is exergy completely destroyed?
8.38 Give the exergy balance for a closed system.
8.39 Explain the statement: The exergy of an isolated system can never increase.
How is it related to the principle of increase of entropy?
8.40 Give the exergy balance of a steady flow system.
8.41 Derive expressions for the irreversibility and second law efficiency of a
(i) steam turbine
(ii) compressor
(iii) heat exchanger, and
(iv) mixer
8.42 What is the deficiency of the first law efficiency? How does the second law
efficiency make up this deficiency?
8.43 How can you improve the first law efficiency and the second law efficiency?
Problems
8.1 What is the maximum useful work which can be obtained when 100 kJ are
abstracted from a heat reservoir at 675 K in an environment at 288 K? What is
the loss of useful work if (a) a temperature drop of 50°C is introduced between
the heat source and the heat engine, on the one hand, and the heat engine and
the heat sink, on the other, (b) the source temperature drops by 50°C and the
sink temperature rises by 50°C during the heat transfer process according to
d-Q
= ± constant?
Ans. (a) 11.2 kJ, (b) 5.25 kJ
dT
8.2 In a steam generator, water is evaporated at 260°C, while the combustion gas
(cp = 1.08 kJ/kg K) is cooled from 1300°C to 320°C. The surroundings are
at 30°C. Determine the loss in available energy due to the above heat transfer
per kg of water evaporated. (Latent heat of vaporization of water at 260°C =
1662.5 kJ/kg.)
Ans. 443.6 kJ
8.3 Exhaust gases leave an internal combustion engine at 800°C and 1 atm, after
having done 1050 kJ of work per kg of gas in the engine (cp of gas = 1.1 kJ/kg
K). The temperature of the surroundings is 30°C. (a) How much available
energy per kg of gas is lost by throwing away the exhaust gases? (b) What is
the ratio of the lost available energy to the engine work?
Ans. (a) 425.58 kJ, (b) 0.405
8.4 A hot spring produces water at a temperature of 56°C. The water flows into a
large lake, with a mean temperature of 14°C, at a rate of 0.1 m3 of water per min.
What is the rate of working of an ideal heat engine which uses all the available
energy?
Ans. 19.5 kW
8.5 0.2 kg of air at 300°C is heated reversibly at constant pressure to 2066 K. Find
the available and unavailable energies of the heat added. Take T0 = 30°C and
cp = 1.0047 kJ/kg K.
Ans. 211.9 and 78.1 kJ
the linear law
268
Engineering Thermodynamics
8.6 Eighty kg of water at 100°C are mixed with 50 kg of water at 60°C, while the
temperature of the surroundings is 15°C. Determine the decrease in available
energy due to mixing.
Ans. 236 kJ
8.7 A lead storage battery used in an automobile is able to deliver 5.2 MJ of
electrical energy. This energy is available for starting the car.
Let compressed air be considered for doing an equivalent amount of work in
starting the car. The compressed air is to be stored at 7 MPa, 25°C. What is the
volume of the tank that would be required to let the compressed air have an
availability of 5.2 MJ? For air, pv = 0.287 T, where T is in K, p in kPa, and v in
m3 /kg.
Ans. 0.228 m3
8.8 Ice is to be made from water supplied at 15°C by the process shown in Fig.
8.33. The final temperature of the ice is – 10°C, and the final temperature of the
water that is used as cooling water in the condenser is 30°C. Determine the
minimum work required to produce 1000 kg of ice.
Ans. 33.37 MJ
Take cp for water = 4.187 kJ/kg K, cp for ice = 2.093 kJ/kg K, and latent heat of
fusion of ice = 334 kJ/kg.
Water 30°C
Q1
Water 15°C
Wrev
R
Q2
Ice, – 10°C
Fig. 8.33
8.9 A pressure vessel has a volume of 1 m3 and contains air at 1.4 MPa, 175°C.
The air is cooled to 25°C by heat transfer to the surroundings at 25°C. Calculate the availability in the initial and final states and the irreversibility of this
process.
Take p0 = 100 kPa.
Ans. 135 kJ/kg, 114.6 kJ/kg, 222 kJ
8.10 Air flows through an adiabatic compressor at 2 kg/s. The inlet conditions are
1 bar and 310 K and the exit conditions are 7 bar and 560 K. Compute the net
rate of availability transfer and the irreversibility. Take T0 = 298 K.
Ans. 481.1 kW and 21.2 kW
8.11 An adiabatic turbine receives a gas (cp = 1.09 and cv = 0.838 kJ/kg K) at 7 bar
and 1000°C and discharges at 1.5 bar and 665°C. Determine the second law
and isentropic efficiencies of the turbine. Take T0 = 298 K. Ans. 0.956, 0.879
8.12 Air enters an adiabatic compressor at atmospheric conditions of 1 bar, 15°C
and leaves at 5.5 bar. The mass flow rate is 0.01 kg/s and the efficiency of the
compressor is 75%. After leaving the compressor, the air is cooled to 40°C in
an aftercooler. Calculate (a) the power required to drive the compressor, and
(b) the rate of irreversibility for the overall process (compressor and cooler).
Ans. (a) 2.42 kW, (b) 1 kW
8.13 In a rotary compressor, air enters at 1.1 bar, 21°C where it is compressed
adiabatically to 6.6 bar, 250°C. Calculate the irreversibility and the entropy
Available Energy, Availability and Irreversibility
269
production for unit mass flow rate. The atmosphere is at 1.03 bar, 20°C. Neglect the K.E. changes.
Ans. 19 kJ/kg, 0.064 kJ/kg K
8.14 In a steam boiler, the hot gases from a fire transfer heat to water which vaporizes at a constant temperature of 242.6°C (3.5 MPa). The gases are cooled from
1100 to 430°C and have an average specific heat, cp = 1.046 kJ/kg K over this
temperature range. The latent heat of vaporization of steam at 3.5 MPa is
1753.7 kJ/kg. If the steam generation rate is 12.6 kg/s and there is negligible
heat loss from the boiler, calculate: (a) the rate of heat transfer, (b) the rate of
loss of exergy of the gas, (c) the rate of gain of exergy of the steam, and (d) the
rate of entropy generation. Take T0 = 21°C.
Ans. (a) 22096 kW, (b) 15605.4 kW (c) 9501.0 kW, (d) 20.76 kW/K
8.15 An economizer, a gas-to-water finned tube heat exchanger, receives 67.5 kg/s
of gas, cp = 1.0046 kJ/kg K, and 51.1 kg/s of water, cp = 4.186 kJ/kg K. The water
rises in temperature from 402 to 469 K, where the gas falls in temperature from
682 K to 470 K. There are no changes of kinetic energy, and p0 = 1.03 bar and
T0 = 289 K. Determine: (a) rate of change of availability of the water, (b) the rate
of change of availability of the gas, and (c) the rate of entropy generation.
Ans. (a) 4802.2 kW, (b) 7079.8 kW, (c) 7.73 kW/K
8.16 The exhaust gases from a gas turbine are used to heat water in an adiabatic
counterflow heat exchanger. The gases are cooled from 260 to 120°C, while
water enters at 65°C. The flow rates of the gas and water are 0.38 kg/s and
0.50 kg/s respectively. The constant pressure specific heats for the gas and
water are 1.09 and 4.186 kJ/kg K respectively. Calculate the rate of exergy loss
due to heat transfer. Take T0 = 35°C.
Ans. 12.5 kW
8.17 The exhaust from a gas turbine at 1.12 bar, 800 K flows steadily into a heat
exchanger which cools the gas to 700 K without significant pressure drop.
The heat transfer from the gas heats an air flow at constant pressure, which
enters the heat exchanger at 470 K. The mass flow rate of air is twice that of the
gas and the surroundings are at 1.03 bar, 20°C. Determine: (a) the decrease in
availability of the exhaust gases, and (b) the total entropy production per kg
of gas. (c) What arrangement would be necessary to make the heat transfer
reversible and how much would this increase the power output of the plant
per kg of turbine gas? Take cp for exhaust gas as 1.08 and for air as 1.05 kJ/kg
K. Neglect heat transfer to the surroundings and the changes in kinetic and
potential energy.
Ans. (a) 66 kJ/kg, (b) 0.0731 kJ/kg K, (c) 38.7 kJ/kg
8.18 An air preheater is used to heat up the air used for combustion by cooling the
outgoing products of combustion from a furnace. The rate of flow of the
products is 10 kg/s, and the products are cooled from 300°C to 200°C, and for
the products at this temperature cp = 1.09 kJ/kg K. The rate of air flow is
9 kg/s, the initial air temperature is 40°C, and for the air cp = 1.005 kJ/kg K.
(a) What is the initial and final availability of the products?
(b) What is the irreversibility for this process?
(c) If the heat transfer from the products were to take place reversibly
through heat engines, what would be the final temperature of the air?
What power would be developed by the heat engines? Take T0 = 300 K.
Ans. (a) 85.97, 39.68 kJ/kg, (b) 256.5 kW, (c) 394.41 K, 353.65 kW
8.19 A mass of 2 kg of air in a vessel expands from 3 bar, 70°C to 1 bar, 40°C, while
receiving 1.2 kJ of heat from a reservoir at 120°C. The environment is at 0.98
bar, 27°C. Calculate the maximum work and the work done on the atmosphere.
Ans. 177 kJ, 112.5 kJ
270
Engineering Thermodynamics
8.20 Air enters the compressor of a gas turbine at 1 bar, 30°C and leaves the
compressor at 4 bar. The compressor has an efficiency of 82%. Calculate per
kg of air (a) the work of compression, (b) the reversible work of compression,
and (c) the irreversibility. For air, use
T2 s
T1
=
Fp I
GH p JK
g - 1/ g
2
1
where T2s is the temperature of air after isentropic compression and
g = 1.4. The compressor efficiency is defined as (T2s – T1)/(T2 – T1), where T2
is the actual temperature of air after compression.
Ans. (a) 180.5 kJ/kg, (b) 159.5 kJ/kg (c) 21 kJ/kg
8.21 A mass of 6.98 kg of air is in a vessel at 200 kPa, 27°C. Heat is transferred to the
air from a reservoir at 727°C. until the temperature of air rises to 327°C. The
environment is at 100 kPa, 17°C. Determine (a) the initial and final availability
of air, (b) the maximum useful work associated with the process.
Ans. (a) 103.5, 621.9 kJ (b) 582 kJ
8.22 Air enters a compressor in steady flow at 140 kPa, 17°C and 70 m/s and leaves
it at 350 kPa, 127°C and 110 m/s. The environment is at 100 kPa, 7°C. Calculate
per kg of air (a) the actual amount of work required, (b) the minimum work
required, and (c) the irreversibility of the process.
Ans. (a) 114.4 kJ, (b) 97.3 kJ, (c) 17.1 kJ
8.23 Air expands in a turbine adiabatically from 500 kPa, 400 K and 150 m/s to
100 kPa, 300 K and 70 m/s. The environment is at 100 kPa, 17°C. Calculate per
kg of air (a) the maximum work output, (b) the actual work output, and (c) the
irreversibility.
Ans. (a) 159 kJ, (b) 109 kJ, (c) 50 kJ
8.24 Calculate the specific exergy of air for a state at 2 bar, 393.15 K when
the surroundings are at 1 bar, 293.15 K. Take cp = 1 and R = 0.287 kJ/kg K.
Ans. 72.31 kJ/kg
8.25 Calculate the specific exergy of CO2 (cp = 0.8659 and R = 0.1889 kJ/kg K) for a
state at 0.7 bar, 268.15 K and for the environment at 1.0 bar and 293.15 K.
Ans. – 18.77 kJ/kg
8.26 A pipe carries a stream of brine with a mass flow rate of 5 kg/s. Because of
poor thermal insulation the brine temperature increases from 250 K at the pipe
inlet to 253 K at the exit. Neglecting pressure losses, calculate the irreversibility rate (or rate of energy degradation) associated with the heat leakage. Take
Ans. 7.05 kW
T0 = 293 K and cp = 2.85 kJ/kg K.
8.27 In an adiabatic throttling process, energy per unit mass of enthalpy remains
∑
the same. However, there is a loss of exergy. An ideal gas flowing at the rate m
is throttled from pressure p1 to pressure p 2 when the environment is at tem
perature T0. What is the rate of exergy loss due to throttling?
∑
∑
Ans. I = m RT 0 ln
p1
p2
8.28 Air at 5 bar and 20°C flows into an evacuated tank until the pressure in the
tank is 5 bar. Assume that the process is adiabatic and the temperature of the
surroundings is 20°C. (a) What is the final temperature of the air? (b) What is
the reversible work produced between the initial and final states of the air? (c)
What is the net entropy change of the air entering the tank? (d) Calculate the
irreversibility of the process.
Ans. (a) 410.2 K, (b) 98.9 kJ/kg, (c) 0.3376 kJ/kg K, (d) 98.9 kJ/kg
Available Energy, Availability and Irreversibility
271
8.29 A Carnot cycle engine receives and rejects heat with a 20°C temperature
differential between itself and the thermal energy reservoirs. The expansion
and compression processes have a pressure ratio of 50. For 1 kg of air as the
working substance, cycle temperature limits of 1000 K and 300 K and T0 =
280 K, determine the second law efficiency.
Ans. 0.965
8.30 Energy is received by a solar collector at the rate of 300 kW from a source
temperature of 2400 K. If 60 kW of this energy is lost to the surroundings at
steady state and if the user temperature remains constant at 600 K, what are
the first law and the second law efficiencies? Take T0 = 300 K.
Ans. 0.80, 0.457
8.31 For flow of an ideal gas through an insulated pipeline, the pressure drops from
100 bar to 95 bar. If the gas flows at the rate of 1.5 kg/s and has cp = 1.005 and
cv = 0.718 kJ/kg-K and if T0 = 300 K, find the rate of entropy generation, and
rate of loss of exergy.
Ans. 0.0215 kW/K, 6.46 kW
8.32 The cylinder of an internal combustion engine contains gases at 2500°C, 58 bar.
Expansion takes place through a volume ratio of 9 according to pv1.38 = const.
The surroundings are at 20°C, 1.1 bar. Determine the loss of availability, the
work transfer and the heat transfer per unit mass. Treat the gases as ideal
having R = 0.26 kJ/kg-K and cv = 0.82 kJ/kg-K.
Ans. 1144 kJ/kg, 1074 kJ/kg, – 213 kJ/kg
8.33 In a counterflow heat exchanger, oil (cp = 2.1 kJ/kg-K) is cooled from 440 to
320 K, while water (cp = 4.2 kJ/kg K) is heated from 290 K to temperature T. The
respective mass flow rates of oil and water are 800 and 3200 kg/h. Neglecting
pressure drop, KE and PE effects and heat loss, determine (a) the temperature
T, (b) the rate of exergy destruction, (c) the second law efficiency. Take T0 =
17°C and p0 = 1 atm.
Ans. (a) 305 K, (b) 41.4 MJ/h, (c) 10.9%
8.34 Oxygen enters a nozzle operating at steady state at 3.8 MPa, 387°C and
10 m/s. At the nozzle exit the conditions are 150 kPa, 37°C and 750 m/s. Determine (a) the heat transfer per kg and (b) the irreversibility. Assume oxygen as
an ideal gas, and take T0 = 20°C, p0 = 1 atm.
Ans. (a) – 37.06 kJ/kg, (b) 81.72 kJ/kg
8.35 Argon gas expands adiabatically in a turbine from 2 MPa, 1000°C to 350 kPa.
The mass flow rate is 0.5 kg/s and the turbine develops power at the rate of
120 kW. Determine (a) the temperature of argon at the turbine exit, (b) the
irreversibility rate, and (c) the second law efficiency. Neglect KE and PE effects and take T0 = 20°C, p0 = 1 atm.
Ans. (a) 538.1°C, b) 18.78 kW, (c) 86.5%
8.36 In the boiler of a power plant are tubes through which water flows as it is
brought from 0.8 MPa, 150°C (h = 632.6 kJ/kg, s = 1.8418 kJ/kg K) to 0.8 MPa,
250°C (h = 2950 kJ/kg, s = 7.0384 kJ/kg K). Combustion gases passing over the
tubes cool from 1067°C to 547°C. These gases may be considered as air (ideal
gas) having cp = 1.005 kJ/kg K. Assuming steady state and neglecting any
heat loss, and KE and PE effects, determine (a) the mass flow rate of combustion gases per kg of steam, (b) the loss of exergy per kg steam, and (c) the
second law efficiency. Take T0 = 25°C, p0 = 1 atm.
Ans. (a) mg /mw = 4.434, (b) 802.29 kJ/kg steam, (c) 48.9%
8.37 Air enters a hair dryer at 22°C, 1 bar with a velocity of 3.7 m/s and exits at 83°C,
1 bar with a velocity of 9.1 m/s through an area of 18.7 cm2. Neglecting any heat
272
Engineering Thermodynamics
loss and PE effect and taking T0 = 22°C, (a) evaluate the power required in kW,
and (b) devise and evaluate a second law efficiency.
Ans. (a) – 1.02 kW,
(b) 9%
8.38 An isolated system consists of two solid blocks. One block has a mass of
5 kg and is initially at 300°C. The other block has a mass of 10 kg and is initially
at – 50°C. The blocks are allowed to come into thermal equilibrium. Assuming
the blocks are incompressible with constant specific heats of 1 and 0.4 kJ/kg
K, respectively, determine (a) the final temperature, (b) the irreversibility. Take
T0 = 300 K.
Ans. (a) 417.4 K, (b) 277 kJ
8.39 Air flows into a heat engine at ambient conditions 100 kPa, 300 K. Energy is
supplied as 1200 kJ per kg air from a 1500 K source and in some part of the
process, a heat loss of 300 kJ/kg air happens at 750 K. The air leaves the
engine at 100 kPa, 800 K. Find the first and the second law efficiencies.
Ans. 0.315, 0.672
8.40 Consider two rigid containers each of volume 1 m3 containing air at 100 kPa,
400 K. An internally reversible Carnot heat pump is then thermally connected
between them so that it heats one up and cools the other down. In order to
transfer heat at a reasonable rate, the temperature difference between the
working fluid inside the heat pump and the air in the containers is set to 20°C.
The process stops when the air in the coldest tank reaches 300 K. Find the
final temperature of the air that is heated up, the work input to the heat pump,
and the overall second law efficiency.
Ans. 550 K, 31.2 kJ, 0.816
8.41 Argon enters an insulated turbine operating at steady state at 1000°C and
2 MPa and exhausts at 350 kPa. The mass flow rate is 0.5 kg/s and the turbine
develops power at the rate of 120 kW. Determine (a) the temperature of the
argon at the turbine exit, (b) the irreversibility of the turbine, (c) the second
law efficiency. Neglect KE and PE effects. Take T0 = 20°C, p0 = 1 bar.
Ans. (a) 812 K, (b) 18.87 kW, (c) 86.4%
9
Properties of Pure
Substances
A pure substance is a substance of constant chemical composition throughout its
mass. It is a one-component system. It may exist in one or more phases. Let us take
water as the representative of a pure substance. We will study the behaviour of
water in all the three phases in thermodynamic plots on p–v, p–T, T–s and h–s
coordinates.
9.1 P-V DIAGRAM FOR A PURE SUBSTANCE
Assume a unit mass of ice (solid water) at – 10°C and 1 atm contained in a cylinder
and piston machine (Fig. 9.1). Let the ice be heated slowly so that its temperature is
always uniform. The changes which occur
1 atm
in the mass of water would be traced as the
temperature is increased while the pressure
is held constant. Let the state changes of
H2O
Q
water be plotted on p–v co-ordinates. The
distinct regimes of heating, as shown in
Fig. 9.2, are:
1–2 The temperature of ice increases
from –10°C to 0°C. The volume of ice Fig. 9.1 Heating of H O at a
2
would increase, as would be the case for
Constant Pressure of 1 Atm
any solid upon heating. At state 2, i.e. 0°C,
the ice would start melting.
2–3 Ice melts into water at a constant temperature of 0°C. At state 3, the melting
process ends. There is a decrease in volume, which is a peculiarity of water.
3–4 The temperature of water increases, upon heating, from 0°C to 100°C. The
volume of water increases because of thermal expansion.
4–5 The water starts boiling at state 4 and boiling ends at state 5. This phase
change from liquid to vapour occurs at a constant temperature of 100°C (the
pressure being constant at 1 atm). There is a large increase in volume.
5–6 The vapour is heated to, say, 250°C (state 6). The volume of vapour
increases from v5 to v6 .
274
Engineering Thermodynamics
p
2 atm
1 atm
0.5 atm
3 4
1
2
5
6
Large volume change
3 4
1
2
5
6
3 4
1
2
5
6
V
Fig. 9.2
Changes in the Volume of Water During Heating at Constant
Pressure
Water existed in the solid phase between 1 and 2, in the liquid phase between 3 and
4, and in the gas phase beyond 5. Between 2 and 3, the solid changed into the liquid
phase by absorbing the latent heat of fusion and between 4 and 5, the liquid changed
into the vapour phase by absorbing the latent heat of vaporization, both at constant
temperature and pressure.
The states 2, 3, 4 and 5 are known as saturation states. A saturation state is a
state from which a change of phase may occur without a change of pressure or
temperature. State 2 is a saturated solid state because a solid can change into liquid
at constant pressure and temperature from state 2. States 3 and 4 are both saturated
liquid states. In state 3, the liquid is saturated with respect to solidification, whereas
in state 4, the liquid is saturated with respect to vaporization. State 5 is a saturated
vapour state, because from state 5, the vapour can condense into liquid without a
change of pressure or temperature.
If the heating of ice at – 10°C to steam at 250°C were done at a constant pressure
of 2 atm, similar regimes of heating would have been obtained with similar
saturation states 2, 3, 4 and 5, as shown in Fig. 9.2. All the state changes of the
system can similarly be plotted on the p-v coordinates, when it is heated at different
constant pressures. All the saturated solid states 2 at various pressures are joined by
a line, as shown in Fig. 9.3.
Sat. solid line
Critical state
p
S
+
3 L 2 4
3 2 4
3 4 2
1
3 4
5
5
V
Saturated vapour line
5
2
5
Saturated
liquid lines
6
Triple point line
L+V
S
S+V
V
Fig. 9.3
p-v Diagram of Water, Whose Volume Decreases on Melting
275
Properties of Pure Substances
Similarly, all the saturated liquid states 3 with respect to solidification, all the
saturated liquid states 4 with respect to vaporization, and all the saturated vapour
states 5, are joined together.
Figure 9.4 shows state changes of a pure substance other than water whose
volume increases on melting.
Saturated liquid lines
p
Critical state
2
3
2
3
2
5
4
3
2
1
L
4
3
S
4 L+V
4
V
Saturated vapour line
5
5
6
5
Triple point line
L
Saturated
solid line
S
V
S+V
V
Fig. 9.4
p-v Diagram of a Pure Substance Other Than Water, Whose
Volume Increases on Melting
The line passing through all the saturated solid states 2 (Figs. 9.3 and 9.4) is called
the saturated solid line. The lines passing through all the saturated liquid states 3
and 4 with respect to solidification and vaporization respectively are known as the
saturated liquid lines, and the line passing through all the saturated vapour states 5,
is the saturated vapour line. The saturated liquid line with respect to vaporization
and the saturated vapour line incline towards each other and form what is known as
the saturation or vapour dome. The two lines meet at the critical state.
To the left of the saturated solid line is the solid (S) region (Fig. 9.4). Between
the saturated solid line and saturated liquid line with respect to solidification there
exists the solid-liquid mixture (S + L) region. Between the two saturated liquid
lines is the compressed liquid region. The liquid-vapour mixture region (L + V )
exists within the vapour dome between the saturated liquid and saturated vapour
lines. to the right of the saturated vapour line is the vapour region. The triple point
is a line on the p-v diagram, where all the three phases, solid, liquid, and gas, exist
in equilibrium. At a pressure below the triple point line, the substance cannot exist
in the liquid phase, and the substance, when heated, transforms from solid to vapour
(known as sublimation) by absorbing the latent heat of sublimation from the
surroundings. The region below the triple point line is, therefore, the solid-vapour
(S + V) mixture region. Table 9.1 gives the triple point data for a number of
substances.
276
Table 9.1
Engineering Thermodynamics
Triple-Point Data
Substance
Temperature, K
Acetylene, C2H 2
Ammonia, NH3
Argon, A
Carbon dioxide, CO2
Carbon monoxide, CO
Ethane, C2H 6
Enthylene, C2H4
Hydrogen, H2
Methane, CH4
Nitrogen, N2
Oxygen, O2
Water, H2O
192.4
195.42
83.78
216.55
68.14
89.88
104.00
13.84
90.67
63.15
54.35
273.16
Pressure, mm Hg
962
45.58
515.7
3885.1
115.14
0.006
0.9
52.8
87.7
94.01
1.14
4.587
Liquid is, most often, the working fluid in power cycles, etc. and interest is often
confined to the liquid-vapour regions only. So to locate the state points, the solid
regions from Figs. 9.3 and 9.4 can be omitted. The p-v diagram then becomes as
shown in Fig. 9.5. If the vapour at state A is compressed slowly and isothermally,
the pressure will rise until there is saturated vapour at point B. If the compression is
continued, condensation takes place, the pressure remaining constant so long as the
temperature remains constant. At any point between B and C, the liquid and vapour
are in equilibrium. Since a very large increase in pressure is needed to compress the
liquid, line CD is almost vertical. ABCD is a typical isotherm of a pure substance
on a p-v diagram. Some isotherms are shown in Fig. 9.5. As the temperature
increases, the liquid-vapour transition, as represented by BC, decreases, and
becomes zero at the critical point. Below the critical point only, there is a liquidvapour transition zone, where a saturated liquid, on heating, absorbs the latent heat
of vaporization, and becomes saturated vapour at a constant pressure and
temperature. Similarly, a saturated vapour, on cooling, releases the latent heat of
condensation at constant pressure and temperature to become saturated liquid.
Above the critical point, however, a liquid, upon heating, suddenly flashes into
vapour, or a vapour, upon cooling, suddenly condenses into liquid. There is no
distinct transition zone from liquid to vapour and vice versa. The isotherm passing
through the critical point is called the critical isotherm, and the corresponding
temperature is known as the critical temperature (tc). The pressure and volume at
the critical point are known as the critical pressure (pc ) and the critical volume (vc)
respectively. For water
pc = 221.2 bar
tc = 374.15°C
vc = 0.00317 m3/kg
Properties of Pure Substances
277
Critical point
Pc
pc = 221.2 bar
tc = 374.15°C
vc = 0.00317 m3/kg
Gas phase
D
p
tc
Vapour phase
T increasing
L
C
B
A
V
L+V
Saturated
vapour line
Saturated liquid line
Vc
Fig 9.5
V
Saturation Curve on p-v Diagram
p
The critical point data of certain substances are given in Appendix F. Above the
critical point, the isotherms are continuous curves that at large volumes and low
pressures approach equilateral hyperbolas.
When a liquid or solid is in
Vapour
Vapour
equilibrium with its vapour at a
pressure
given temperature, the vapour Temperature
T
exerts a pressure that depends
Solid or liquid
only on the temperature (Fig. 9.6).
In general, the greater the
temperature, the higher is the
Fig. 9.6 Vapour Pressure
vapour pressure. The temperature
Critical point
at which the vapour pressure is
equal to 760 mm Hg is called the
normal boiling point.
Phase change occurs at
V
L
constant pressure and temA
B
p1
perature. A pure liquid at a given
A
pressure will transform into
(tsat)1
(psat)2
B
L+V
vapour only at a particular
t2
temperature, known as saturation
vfg
temperature, which is a function
vg
vf
v
of pressure. Similarly, if the
temperature is fixed, the liquid
Fig. 9.7 Saturation Pressure and
will boil (or condense) only at a
Temperature
particular pressure, called the
saturation pressure, which is a function of temperature. In Fig. 9.7, if p1 is the
pressure, the corres-ponding saturation temperature is (t sat)1, or if t2 is the given
278
Engineering Thermodynamics
temperature, the saturation pressure is (psat)2 . As the pressure increases, the
saturation temperature increases. Saturation states exist up to the critical point. At
point A, the liquid starts boiling, and at point B, the boiling gets completed. At A, it
is all liquid (saturated) and there is no vapour, while at B, it is all vapour (saturated)
and there is no liquid. Vapour content progressively increases as the liquid changes
its state from A towards B.
If vf is the specific volume of the saturated liquid at a given pressure, and vg the
specific volume of the saturated vapour, then (vg – vf) or vfg is the change in specific
volume during phase transition (boiling or condensation) at the pressure. As
pressure increases, vfg decreases, and at the critical point vfg becomes zero.
9.2 P-T DIAGRAM FOR A PURE SUBSTANCE
The state changes of a pure substance, upon slow heating at different constant
pressures, are shown on the p-v plane, in Figs. 9.2, 9.3, and 9.4. If these state
changes are plotted on p-T coordinates, the diagram, as shown in Fig. 9.8, will be
obtained. If the heating of ice at – 10°C to steam at 250°C at the constant pressure
of 1 atm is considered, 1–2 is the solid (ice) heating, 2–3 is the melting of ice at
0°C, 3–4 is the liquid heating, 4–5 is the vaporization of water at 100°C, and
5–6 is the heating in the vapour phase. The process will be reversed from state 6 to
state 1 upon cooling. The curve passing through the 2, 3 points is called the fusion
curve, and the curve passing through the 4, 5 points (which indicate the vaporization
or condensation at different temperatures and pressures) is called the vaporization
curve. If the vapour pressure of a solid is measured at different temperatures, and
these are plotted, the sublimation curve will be obtained. The fusion curve, the
vaporization curve, and the sublimation curve meet at the triple point.
Vaporization
curve
Fusion
curve
Critical point
p
1
2, 3
1 atm 1
2, 3 Liquid
1
2, 3 Region
Sublimation
curve
6
4, 5
6
6
4, 5
Solid
region
–10°C
4, 5
Triple
point
0°C
Vapour
region
100°C
250°C
T
Fig. 9.8
Phase Equilibrium Diagram on p-T Coordinates
The slopes of the sublimation and vaporization curves for all substances are
positive. The slope of the fusion curve for most substances is positive, but for water,
it is negative. The temperature at which a liquid boils is very sensitive to pressure,
as indicated by the vaporization curve which gives the saturation temperatures at
different pressures, but the temperature at which a solid melts is not such a strong
function of pressure, as indicated by the small slope of the fusion curve.
279
Properties of Pure Substances
The triple point of water is at 4.58 mm Hg and 273.16 K, whereas that of CO2 is
at 3885 mm Hg (about 5 atm) and 216.55 K. So when solid CO 2 (‘dry ice’) is
exposed to 1 atm pressure, it gets transformed into vapour directly, absorbing the
latent heat of sublimation from the surroundings, which gets cooled or
‘refrigerated’.
9.3 P-V-T SURFACE
The relationships between pressure, specific volume, and temperature can be clearly
understood with the aid of a three-dimensional p-v-T surface. Figure 9.9 illustrates
a substance like water that expands upon freezing and Fig. 9.10 illustrates
substances other than water which contract upon freezing. The projections on the
p-T and p-v planes are also shown in these figures.
Liquid
Pressure
Critical
point
Li
va qui
Tri po dple ur
line
Solid
So
Sp
ec
ific
Va
p
ou
r
lid
vo
-va
lum
po
ur
e
pe
Tem
Tc
u
rat
re
Solid
(a)
L
Solid
Liquid
Critical
point
L
V
S
Vapour
Pressure
Pressure
S
Critical
point
Liquid
vapour
Vap
ou
Triple point
Triple line
Solid-vapour
Temperature
Specific volume
V
(b)
Fig. 9.9
r
T > Tc
Tc
T < Tc
(c)
p-v-T Surface and Projections for a Substance that Expands
on Freezing. (a) Three-dimensional View. (b) Phase Diagram.
(c) p-v Diagram.
280
Solid
ou
lid
ec
ific
vo
Tc
r
So
Constantpressure line
p
Va
Liq
va uidpo
ur
Tri
ple
line
Sp
Critical
point
Liquid
Solid-liq
Pressure
uid
Engineering Thermodynamics
-va
po
ur
e
e
tur
era
p
Tem
lum
S
Solid
Solid-liquid
(a)
L
Critical point
Solid
L
V Vapour
Triple point
Temperature
(b)
Fig. 9.10
Pressure
Pressure
Liquid
Critical
point
Liuidvapour
Triple-line
Solid-vapour
T > Tc
Vapour Tc
T < Tc
Specific-volume
(c)
p-v-T Surface and Projections for a Substance that
Contracts on Freezing. (a) Three-dimensional View.
(b) Phase Diagram. (c) p-v Diagram.
Any point on the p-v-T surface represents an equilibrium state of the substance.
The triple point line when projected to the p-T plane becomes a point. The critical
isotherm has a point of inflection at the critical point.
9.4 T-S DIAGRAM FOR A PURE SUBSTANCE
The heating of the system of 1 kg of ice at – 5°C to steam at 250°C is again
considered, the pressure being maintained constant at 1 atm. The entropy increases
of the system in different regimes of heating are given as follows.
1. The entropy increase of ice as it is heated from – 5°C to 0°C at 1 atm. (cpice =
2.093 kJ/kg K).
281
Properties of Pure Substances
D s1 = s2 – s1 =
z
_
dQ
=
T
= 1 ¥ 2.093 ln
z
T2 = 273 mc p d T
T1 = 268
T
= mcp ln
273
268
273
= 0.0398 kJ/kg K
268
2. The entropy increase of ice as it melts into water at 0°C (latent heat of fusion
of ice = 334.96 kJ/kg)
D s2 = s3 – s2 =
334.96
= 1.23 kJ/kg K
273
3. The entropy increase of water as it is heated from 0°C to 100°C (cp
=
water
4.187 kJ/kg K)
D s3 = s4 – s3 = mcp ln
T3
373
= 1 ¥ 4.187 ln
= 1.305 kJ/kg K
273
T2
4. The entropy increase of water as it is vaporized at 100°C, absorbing the latent
heat of vaporization (2257 kJ/kg)
D s4 = s5 – s4 =
2257
= 6.05 kJ/kg K
273
5. The entropy increase of vapour as it is heated from 100°C to 250°C at
1 atm
D s5 = s6 – s5 =
z
523
373
mc p
dT
523
= 1 ¥ 2.093 ln
373
T
= 0.706 kJ/kg K
assuming the average specific heat of steam in the temperature range of 100°C to
250°C as 2.093 kJ/kg K.
These entropy changes are shown in Fig. 9.11. The curve 1-2-3-4-5-6 is the
isobar of 1 atm. If, during the heating process, the pressure had been maintained
constant at 2 atm, a similar curve would be obtained. The states 2, 3, 4, and 5 are
saturation states. If these states for different pressures are joined, as in Figs. 9.3 and
9.4, the phase equilibrium diagram of a pure substance on the T-s coordinates, as
shown in Fig. 9.12, would be obtained.
Most often, liquid-vapour transformations only are of interest, and Fig. 9.13
shows the liquid, the vapour, and the transition zones only. At a particular pressure,
sf is the specific entropy of saturated water, and sg is that of saturated vapour. The
entropy change of the system during the phase change from liquid to vapour at that
pressure is sfg (= sg – sf). The value of sfg decreases as the pressure increases, and
becomes zero at the critical point.
282
Engineering Thermodynamics
250 ∞C
6
100 ∞C
2
0 °C
1
1
- 5 ∞C
4
2 atm
5
4
1 atm
5
6
3
2
3
Ds2
Ds1
s1 s2
Ds3
s3
Ds4
Ds5
s4
s5
s6
s
Fig. 9.11
Isobars on T-s Plot
Critical state
p=c
L
S+L
4
6
5
p=c
2
5
4
3
S
V
Triple point line
L+V
2
3
1
S+V
s
Fig. 9.12 Phase Equilibrium Diagram on T-s Coordinates
9.5 h-s DIAGRAM OR MOLLIER DIAGRAM FOR A PURE
SUBSTANCE
From the first and second laws of thermodynamics, the following property relation
was obtained.
Tds = dh – vdp
or
FG ∂h IJ = T
H ∂s K
(9.1)
p
This equation forms the basis of the h-s diagram of a pure substance, also called
the Mollier diagram. The slope of an isobar on the h-s coordinates is equal to the
absolute saturation temperature (t sat + 273) at that pressure. If the temperature remains constant the slope will remain constant. If the temperature increases, the
slope of the isobar will increase.
283
Properties of Pure Substances
Critical point
tc = 374.15∞C
ba
r
Saturated vapour line
T
p=
c
p0
.2
21
=2
p=c
L
4
5
V
L+V
4
Saturated liquid line 5
4
p increasing
5
sfg
sf
sg
S
Fig. 9.13 Saturation (or vapour) Dome for Water
Consider the heating of a system of ice at – 5°C to steam at 250°C, the pressure
being maintained constant at 1 atm. The slope of the isobar of 1 atm on the h-s
coordinates (Fig. 9.14) first increases as the temperature of the ice increases from –
5°C to 0°C (1–2). Its slope then remains constant as ice melts into water at the
constant temperature of 0°C (2–3). The slope of the isobar again increases as the
temperature of water rises from 0°C to 100°C (3–4). The slope again remains
constant as water vaporizes into steam at the constant temperature of 100°C
(4–5). Finally, the slope of the isobar continues to increase as the temperature of
steam increases to 250°C (5–6) and beyond. Similarly, the isobars of different
pressures can be drawn on the h-s diagram as shown in Figs. 9.14 and 9.15. States
2, 3, 4 and 5 are saturation states. Figure 9.15 shows the phase equilibrium diagram
of a pure substance on the h-s coordinates, indicating the saturated solid line,
saturated liquid lines and saturated vapour line, the various phases, and the transition
(mixture) zones.
1 atm
6
6
Increasing slope
5
h
5
4
1
1
2
3
2
Slope constant
4
3
Increasing slope
Slope constant
Increasing slope
s
Fig. 9.14
Isobars on h-s Plot
284
Engineering Thermodynamics
6
Critical point
6
5
S+L
4
L
4
2
1
Saturated vapour line
L+V
3
S
V
5
S+V
3
Triple point line
2
1
Saturated liquid line
s
Fig. 9.15
Phase Equilibrium Diagram on h-s Coordinates (Mollier diagram)
Figure 9.16 is the h-s or the Mollier diagram indicating only the liquid and
vapour phases. As the pressure increases, the saturation temperature increases, and
so the slope of the isobar also increases. Hence, the constant pressure lines diverge
from one another, and the critical isobar is a tangent at the critical point, as shown.
In the vapour region, the states of equal slopes at various pressures are joined by
lines, as shown, which are the constant temperature lines. Although the slope of an
isobar remains continuous beyond the saturated vapour line, the isotherm bends
towards the right and its slope decreases asymptotically to zero, because in the
ideal gas region it becomes horizontal and the constant enthalpy implies constant
temperature. At a particular pressure, hf is the specific enthalpy of saturated water,
hg is that of saturated vapour, and hfg (= hg – hf) is the latent heat of vaporization at
that pressure. As the pressure increases, hfg decreases, and at the critical pressure,
hfg becomes zero.
pcr = 221.2 bar
Constant
temperature lines
h
Critical
point
Sat liq
line
p=
c
c
p=
c
hg
hfg
t=c
p=
V
p=c
hf
t=c
L+V
Saturated
vapour line
sfg
sf
s
Fig. 9.16
sg
Enthalpy-entropy Diagram of Water
285
Properties of Pure Substances
9.6 QUALITY OR DRYNESS FRACTION
If in 1 kg of liquid-vapour mixture, x kg is the mass of vapour and (1 – x) kg is the
mass of liquid, then x is known as the quality or dryness fraction of the liquid vapour
mixture. Therefore, quality indicates the mass fraction of vapour in a liquid vapour
mixture, or
mv
mv + ml
x=
where mv and ml are the masses of vapour and liquid respectively in the mixture.
The value of x varies between 0 and 1. For saturated water, when water just starts
boiling, x = 0, and for saturated vapour, when vaporization is complete, x = 1, for
which the vapour is said to be dry saturated.
Points m in Fig. 9.17 (a), (b), and (c) indicate the saturated liquid states with
x = 0, and points n indicate the saturated vapour states with x = 1, the lines mn
indicating the transition from liquid to vapour. Points a, b, and c at various pressures
indicate the situations when the masses of vapour reached 25%, 50%, and 75% of
the total mass, i.e. at points a, the mass of liquid is 75% and the mass of vapour is
25% of the total mass, at points b, the mixture consists of 50% liquid and 50%
vapour by mass, and at points c, the mixture consists of 75% vapour and 25% liquid
by mass. The lines passing through points a, b and c are the constant quality lines of
0.25, 0.50, and 0.75 respectively. Constant quality lines start from the critical point.
Critical point
a
m a
x=0
n
m a b
T
m
a
m
a
m
b n
c
n
b c
n
x = 0.75 x = 1.0
x = 0 x = 0.25 x = 0.50
b
c
a
b
a
b
n
c
n
m
c
m
c
n
x = 0.25 x = 0.50 x = 0.75
s
v
(a)
(b)
Critical point
h
p
Critical point
x=0
sat.
liq
line
n
m
m
m
m
a
a
b
n
c
b
c
Saturated vapour line
n
c
n
b
c
a
x=
b
0.7
a
5
x = 0.50
x = 0.25
x=1
s
(c)
Fig. 9.17 Constant Quality Lines on p-v, T-s and h-s Diagrams
x = 1.0
286
Engineering Thermodynamics
Let V be the total volume of a liquid vapour mixture of quality x, Vf the volume
of the saturated liquid, and Vg the volume of the saturated vapour, the corresponding
masses being m, mf, and mg respectively.
Now
m = mf + mg
and
V = Vf + Vg
mv = mf vf + mg vg
= (m – mg) vf + mg vg
FG
H
\
v = 1-
IJ v + m v
mK
m
mg
g
f
g
v = (1 – x) vf + xvg
where x =
mg
m
(9.2)
, vf = specific volume of saturated liquid, vg = specific volume of
saturated vapour, and v = specific volume of the mixture of quality x.
Similarly
s = (1 – x) sf + xsg
h = (1 – x) hf + xhg
u = (1 – x) uf + xug
(9.3)
(9.4)
(9.5)
where s, h, and u refer to the mixture of quality x, the suffix f and suffix g indicate
the conditions of saturated liquid and saturated vapour respectively. From
Eq. (9.2)
v = (1 – x) vf + xvg
= vf + x(vg – vf)
= vf + x, vfg
or
v = vf + x vfg
(9.6)
h = hf + xhfg
(9.7)
s = sf + xsfg
(9.8)
u = uf + xufg
(9.9)
Similarly
9.7
STEAM TABLES
The properties of water are arranged in the steam tables as functions of pressure
and temperature. Separate tables are provided to give the properties of water in the
saturation states and in the liquid and vapour phases. The internal energy of
saturated water at the triple point (t = 0.01°C) is arbitrarily chosen to be zero.
Since h = u + pv, the enthalpy of saturated water at 0.01°C is slightly positive
because of the small value of (pv) term. The entropy of saturated water is also
chosen to be zero at the triple point.
287
Properties of Pure Substances
9.7.1
Saturation States
When a liquid and its vapour are in equilibrium at a certain pressure and
temperature, only the pressure or the temperature is sufficient to identify the
saturation state. If the pressure is given, the temperature of the mixture gets fixed,
which is known as the saturation temperature, or if the temperature is given, the
saturation pressure gets fixed. Saturated liquid or the saturated vapour has only one
independent variable, i.e. only one property is required to be known to fix up the
state. Tables A.1 (a) and A.1 (b) in the appendix give the properties of saturated
liquid and saturated vapour. In Table A.1(a), the independent variable is
temperature. At a particular temperature, the values of saturation pressure p, and vf,
vg, hf, hfg, hg, sf and sg are given, where vf, hf, and sf refer to the saturated liquid
states; vg, hg and sg refer to the saturated vapour state; and vfg , hfg, and sfg refer to
the changes in the property values during evaporation (or condensation) at that
temperature, where vfg = vg – vf and sfg = sg – sf.
In Table A.1(b), the independent variable is pressure. At a particular pressure,
the values of saturation temperature t, and vf, vg, hf, hfg, hg, sf, and sg are given.
Depending upon whether the pressure or the temperature is given, either Table
A.1(a) or Table A.1(b) can be conveniently used for computing the properties of
saturation states.
If data are required for intermediate temperatures or pressures, linear
interpolation is normally accurate. The reason for the two tables is to reduce the
amount of interpolation required.
9.7.2
Liquid-vapour Mixtures
T
Let us consider a mixture of saturated liquid water and water vapour in equilibrium
at pressure p and temperature t. The composition of the mixture by mass will be
given by its quality x, and its state will be
within the vapour dome (Fig. 9.18). The
properties of the mixture are as given in
Article 9.6, i.e.
t
p
v = vf + xvfg
x
u = uf + xufg
h = hf + xhfg
s = sf + xsfg
sf
s sg
s
Fig. 9.18 Property in Two
Phase Region
where vf, vfg , hf, hfg, sf and sfg are the
saturation properties at the given pressure and temperature.
If p or t and the quality of the mixture are given, the properties of the mixture (v,
u, h and s) can be evaluated from the above equations. Sometimes, instead of quality,
one of the above properties, say, specific volume v, and pressure or temperature are
given. In that case, the quality of the mixture x has to be calculated from the given
v and p or t and then x being known, other properties are evaluated.
288
9.7.3
Engineering Thermodynamics
Superheated Vapour
When the temperature of the vapour is greater than the saturation temperature
corresponding to the given pressure, the vapour is said to be superheated (state 1 in
Fig. 9.19). The difference between the temperature of the superheated vapour and
the saturation temperature at that pressure is called the superheat or the degree of
superheat. As shown in Fig. 9.19, the difference (t1 – tsat) is the superheat.
In a superheated vapour at a
given pressure, the temperature
1
t1
may have different values greater
Subcooling
Superheat
than the saturation temperature.
p
tsat
Table A.2 in the appendix gives
the values of the properties
2
(volume, enthaly, and entropy) of
superheated vapour for each
s
tabulated pair of values of
Fig.
9.19
Superheat
and
Subcooling
pressure and temperature, both of
which are now independent.
Interpolation or extrapolation is to be used for pairs of values of pressure and
temperature not given.
9.7.4
Compressed Liquid
When the temperature of a liquid is less than the saturation temperature at the given
pressure, the liquid is called compressed liquid (state 2 in Fig. 9.19). The pressure
and temperature of compressed liquid may vary independently, and a table of
properties like the superheated vapour table could be arranged, to give the
properties at any p and t. However, the properties of liquids vary little with pressure.
Hence the properties are taken from the saturation tables at the temperature of the
compressed liquid. When a liquid is cooled below its saturation temperature at a
certain pressure it is said to be subcooled. The difference in saturation temperature
and the actual liquid temperature is known as the degree of subcooling, or simply,
subcooling (Fig. 9.19).
9.8 CHARTS OF THERMODYNAMIC PROPERTIES
The presentation of properties of substances in the form of a chart has certain
obvious advantages. The manner of variation of properties is clearly demonstrated
in the chart and there is no problem of interpolation. However, the precision is not
as much as in steam tables.
The temperature-entropy plot and enthalpy-entropy plot (Figs. 9.20 and 9.21)
are commonly used. The temperature-entropy plot shows the vapour dome and the
lines of constant pressure, constant volume, constant enthalpy, constant quality, and
constant superheat. However, its scale is small and limited in use. The enthalpyentropy plot or Mollier chart, has a larger scale to provide data suitable for many
computations. It contains the same data as does the T-s chart. The Mollier chart for
water is also given in Appendix E.I.
289
Properties of Pure Substances
Critical point
P
cr
c
x=
c
T
p=
v=
Compressed
liquid
h=c
Superheated
Vapour
p=c
c
v=
p=c
c
a
b
x=c
h=c
h=c
c
x=c
s
Fig. 9.20 (a) Constant Property Lines on T-s Plot
v=c
p=c
v=c
Critical point
h
t=c
p=c
t=c
p=c
t=c
C.L.
t=c
S.V.
v=c
x=c
x=c
x=c
s
Fig. 9.20 (b) Constant Property Lines on Mollier Diagram
9.9 MEASUREMENT OF STEAM QUALITY
The state of a pure substance gets fixed if two independent properties are given. A pure
substance is thus said to have two degrees of freedom. Of all thermodynamic properties,
it is easiest to measure the pressure and temperature of a substance. Therefore, whenever
pressure and temperature are independent properties, it is the practice to measure them
to determine the state of the substance. This is done in the compressed liquid region or
the superheated vapour region (Fig. 9.22), where the measured values of pressure and
temperature would fix up the state. But when the substance is in the saturation state or
two-phase region (Fig. 9.22), the measured values of pressure and temperature could
290
Engineering Thermodynamics
Fig. 9.21
Mollier Diagram for Steam (Data Taken From Keenan, J.H.,
F.G. Keyes. P.C. Hill and J.G. Moore, Steam Tables, John Wiley,
N.Y., 1969
apply equally well to saturated liquid point f, saturated vapour point g, or to mixtures of
any quality, points x1, x2 or x3. Of the two properties, p and t, only one is independent;
the other is a dependent property. If pressure is given, the saturation temperature gets
automatically fixed for the substance. In order to fix up the state of the mixture, apart
from either pressure or temperature, one more property, such as specific volume,
291
Properties of Pure Substances
p=
c
enthalpy or composition of the mixture (quality) is required to be known. Since it is
relatively difficult to measure the specific volume of a mixture, devices such as
calorimeters are used for determining the quality or the enthalpy of the mixture.
T
t=c
x2
x1
c
t
tsat
g
p=c
=
p
x3
t=c
s
Fig. 9.22
Quality of Liquid-vapour Mixture
In the measurement of quality, the object is always to bring the state of the
substance from the two-phase region to the single-phase or superheated region,
where both pressure and temperature are independent, and measured to fix the state,
either by adiabatic throttling or electric heating.
In the throttling calorimeter, a sample of wet steam of mass m and at pressure p1
is taken from the steam main through a perforated sampling tube (Fig. 9.23). Then
it is throttled by the partially-opened valve (or orifice) to a pressure p2, measured
by mercury manometer, and temperature t2, so that after throttling the steam is in
the superheated region. The process is shown on the T-s and h-s diagrams in Fig.
9.24. The steady flow energy equation gives the enthalpy after throttling as equal to
enthalpy before throttling. The initial and final equilibrium states 1 and 2 are joined
by a dotted line since throttling is irreversible (adiabatic but not isentropic) and the
Steam main
Sampling tube
t2
m
Insulation
Mercury
manometer
p2
p1
p2
Throttle
valve
C.W. out
m
Condenser
Cooling water in
Condensate out
Fig. 9.23
Throttling Calorimeter
292
Engineering Thermodynamics
Critical point
T
h
Critical point
p1
2
h1 =
1
p1
h2
2
p2
t2
t2
1
p2
x1
x1
s
s
(a)
(b)
Fig. 9.24 Throttling Process on T-s and h-s Plot
intermediate states are non-equilibrium states not describable by thermodynamic
coordinates. The initial state (wet) is given by p1 and x1, and the final state by p2
and t2 (superheated). Now
since
h1 = h2
hfp1 + x1hfgp1 = h2
or
x1 =
h2 - h f p1
h fgp1
With p2 and t 2 being known, h2 can be found out from the superheated steam table.
The values of hf and hfg are taken from the saturated steam table corresponding to
pressure p1. Therefore, the quality of the wet steam x1 can be calculated.
To be sure that steam after throttling is in the single-phase or superheated region,
a minimum of 5°C superheat is desired. So if the pressure after throttling is given
and the minimum 5°C superheat is prescribed, then there is the minimum quality of
steam (or the maximum moisture content) at the given pressure p1 which can be
measured by the throttling calorimeter. For example, if p2 = 1 atm., then t2 = 105°C
and the state 2 after throttling gets fixed as shown in Fig. 9.25. From state 2, the
constant enthalpy line intersects the constant pressure p1 line at 1. Therefore, the
quality x1 is the minimum quality that can be measured simply by throttling. If the
h
Critical point
p1
1¢
1
t2 = 105∞C
2¢
t1 = 100∞C
2 ¢¢
1¢¢
x ¢¢1
p2 = 1 atm
2
x 1¢
x1
s
Fig. 9.25 Mnimum Quality That can be Measured only by Throttling
293
Properties of Pure Substances
quality is, say, x ¢1 less than x1, then after throttling to p2 = 1 atm. the superheat after
throttling is less than 5°C. If the quality is x¢¢1, then throttling to 1 atm. does not
give any superheat at all.
When the steam is very wet and the pressure after throttling is not low enough to
take the steam to the superheated region, then a combined separating and throttling
calorimeter is used for the measurement of quality. Steam from the main is first
passed through a separator (Fig. 9.26), where some part of the moisture separates
out due to the sudden change in direction and falls by gravity, and the partially dry
vapour is then throttled and taken to the superheated region. In Fig. 9.27, process
1–2 represents the moisture separation from the wet sample of steam at constant
pressure p1 and process 2–3 represents throttling to pressure p2. With p2 and t3
being measured, h3 can be found out from the superheated steam table.
Stop valve
Throttling valve
p1
p1
Thermometer
m2
t2
Insulation
m
h2
Hg
manometer
m
p2
m1
Sampling
tube
Steam main
C.W. out
Separating
calorimeter
Condenser
Cooling
watar
Separated
moisture (m1)
Fig. 9.26
Condensate (m2)
Separating and Throttling Calorimeter
Now,
h3 = h2 = hfp1 + x2hfgp1
Therefore, x2, the quality of steam after partial moisture separation, can be
evaluated. If m kg of steam is taken through the sampling tube in t secs, m1 kg of it
is separated, and m2 kg is throttled and then condensed to water and collected, then
m = m1 + m2, and at state 2, the mass of dry vapour will be x2 m2. Therefore, the
quality of the sample of steam at state 1, x1 is given by
x1 =
mass of dry vapour at state 1
mass of liquid - vapour mixture at state 1
294
Engineering Thermodynamics
h
Critical point
t3
3
2
p1
1
p2
x1
x2
s
Fig. 9.27
Separating and Throttling Processes on h-s Plot
=
x2 m2
m1 + m2
The quality of wet steam can also be measured by an electric calorimeter
(Fig. 9.28). The sample of steam is passed in steady flow through an electric heater,
as shown. The electrical energy input Q should be sufficient to take the steam to the
superheated region where pressure p2 and temperature t2 are measured. If I is the
current flowing through the heater in amperes and V is the voltage across the coil,
then at steady state Q = VI ¥ 10–3 kW. If m is the mass of steam taken in t seconds
under steady flow condition, then the steady flow energy equation for the heater (as
control volume) gives
w1 h1 + Q = w1 h2
F
H
where w1 is the steam flow rate in kg/s w1 =
Sampling tube
I
K
m
kg/s
t
C.S.
t2
Insulation
Q
p2
p1
I
Steam flow
Steam main
Electric
heater
Exhaust condensed and
collected (m kg in t secs, say)
Fig. 9.28 Electrical Calorimeter
Properties of Pure Substances
295
Q
= h2
w1
With h2, Q and w1 being known, h1 can be computed. Now
\
h1 +
h1 = hfp1 + x1hfgp1
Hence x1 can be evaluated.
Solved Examples
Example 9.1 Find the saturation temperature, the changes in specific volume
and entropy during evaporation, and the latent heat of vaporization of steam at
1 MPa.
Solution
\
\
At 1 MPa, from Table A.1(b) in the Appendix
tsat = 179.91°C
vf = 0.001127 m3/kg
vg = 0.19444 m3/kg
vfg = vg – vf = 0.1933 m3/kg
sf = 2.1387 kJ/kg K
sg = 6.5865 kJ/kg K
sfg = sg – sf = 4.4478 kJ/kg K
hfg = hg – hf = 2015.3 kJ/kg
Example 9.2 Saturated steam has an entropy of 6.76 kJ/kg K. What are its pressure, temperature, specific volume, and enthalpy?
Solution
In Table A.1(b), when sg = 6.76 kJ/kg K
p = 0.6 MPa, t = 158.85°C
vg = 0.3157 m3/kg, and hg = 2756.8 kJ/kg
Example 9.3 Find the enthalpy and entropy of steam when the pressure is
2 MPa and the specific volume is 0.09 m3/kg.
Solution In Table A.1(b), when p = 2 MPa, vf = 0.001177 m3/kg and vg =
0.09963 m3/kg. Since the given volume lies between vf and vg, the substance will be
a mixture of liquid and vapour, and the state will be within the vapour dome. When
in the two-phase region, the composition of the mixture or its quality has to be
evaluated first. Now,
v = vf + x vfg
0.09 = 0.001177 + x (0.09963 – 0.001177)
or
x = 0.904 or 90.4%
At
2 MPa, hf = 908.79 and hfg = 1890.7 kJ/kg
sf = 2.4474 and sfg = 3.8935 kJ/kg K
h = hf + x hfg
= 908.79 + 0.904 ¥ 1890.7 = 2618.79 kJ/kg
s = sf + x sfg
= 2.4474 + 0.904 ¥ 3.8935
= 5.9534 kJ/kg K
296
Example 9.4
380°C.
Engineering Thermodynamics
Find the enthalpy, entropy, and volume of steam at 1.4 MPa,
Solution At p = 1.4 MPa, in Table A.1(b), tsat = 195.07°C. Therefore, the state of
steam must be in the superheated region. In Table A.2, for properties of superheated steam,
at 1.4 MPa, 350°C
v = 0.2003 m3/kg
h = 3149.5 kJ/kg
s = 7.1360 kJ/kg K
and at 1.4 MPa, 400°C v = 0.2178 m3/kg
h = 3257.5 kJ/kg
s = 7.3026 kJ/kg K
\ By interpolation
at 1.4 MPa, 380°C
v = 0.2108 m3/kg
h = 3214.3 kJ/kg
s = 7.2360 kJ/kg K
Example 9.5 A vessel of volume 0.04 m3 contains a mixture of saturated water
and saturated steam at a temperature of 250°C. The mass of the liquid present is
9 kg. Find the pressure, the mass, the specific volume, the enthalpy, the entropy,
and the internal energy.
From Table A.1(a), at 250°C, psat = 3.973 MPa
vf = 0.0012512 m3/kg, vg = 0.05013 m3/kg
hf = 1085.36 kJ/kg,
hfg = 1716.2 kJ/kg
sf = 2.7927 kJ/kg K,
sfg = 3.2802 kJ/kg K
Volume of liquid,
Vf = mf vf
= 9 ¥ 0.0012512 = 0.01126 m3
Volume of vapour,
Vg = 0.04 – 0.01126 = 0.02874 m3
\ Mass of vapour
Solution
mg =
Vg
vg
=
0.02874
= 0.575 kg
0.05013
\ Total mass of mixture,
m = mf + mg = 9 + 0.575 = 9.575 kg
Quality of mixture,
x=
\
mg
m f + mg
=
0.575
= 0.06
9.575
v = vf + xvfs
= 0.0012512 + 0.06 (0.05013 – 0.0012512)
= 0.00418 m3/kg
h = hf + xhfg
= 1085.36 + 0.06 ¥ 1716.2 = 1188.32 kJ/kg
Properties of Pure Substances
297
s = sf + xsfg
= 2.7927 + 0.06 ¥ 3.2802 = 2.9895 kJ/kg K
u = h – pv
= 1188.32 – 3.973 ¥ 103 ¥ 0.00418 = 1171.72 kJ/kg
Also, at 250°C,
uf = 1080.39 and
\
u = uf + xufg
ufg = 1522.0 kJ/kg
= 1080.39 + 0.06 ¥ 1522 = 1071.71 kJ/kg
Example 9.6 Steam initially at 0.3 MPa, 250°C is cooled at constant volume.
(a) At what temperature will the steam become saturated vapour? (b) What is the
quality at 80°C? What is the heat transferred per kg of steam in cooling from
250°C to 80°C?
p
Solution At 0.3 MPa, tsat = 133.55°C
Since t > tsat, the state would be in the su1
0.3 MPa
perheated region (Fig. 9.29). From Table
A.2, for properties of superheated steam,
3
at 0.3 MPa, 250°C
t = 250∞C
80∞C
2
t=?
v1 = 0.7964 m3/kg
t = 80∞C
h1 = 2967.6 kJ/kg
\
v1 = v3 = v2 = 0.7964 m3/kg
v
In Table A.1
Fig. 9.29
when
vg = 0.8919, tsat = 120°C
when
vg = 0.77076, tsat = 125°C
Therefore, when vg = 0.7964, tsat, by
linear interpolation, would be 123.9°. Steam would become saturated vapour at t =
123.9°C
At 80°C,
vf = 0.001029 m3/kg, vg = 3.407 m3/kg,
hf = 334.91 kJ/kg, hfg = 2308.8 kJ/kg, psat = 47.39 kPa
v1 = v2 = 0.7964 m3/kg = vf 80°C + x2vfg80°C
= 0.001029 + x2 (3.407 – 0.001029)
\
x2 =
0.79539
= 0.234
3.40597
h2 = 334.91 + 0.234 ¥ 2308.8 = 875.9 kJ/kg
h1 = 2967.6 kJ/kg
From the first law of thermodynamics
d- Q = du + pdv
\
( d- Q)v = du
or
Q1 – 2 = u2 – u1 = (h2 – p2 v2) – (h1 – p1 v1)
298
Engineering Thermodynamics
= (h2 – h1) + v (p1 – p2)
= (875.9 – 2967.6) + 0.7964 (300 – 47.39)
= – 2091.7 + 201.5 = – 1890.2 kJ/kg
Example 9.7 Steam initially at 1.5 MPa, 300°C expands reversibly and adiabatically in a steam turbine to 40°C. Determine the ideal work output of the turbine per kg of steam.
W
Solution The steady flow energy equation for the control volume, as shown in
Fig. 9.30, gives (other energy terms being neglected)
h1 = h2 + W
\
W = h1 – h2
Work is done by steam at the expense
of a fall in its enthalpy value. The process
is reversible and adiabatic, so it is isentropic. The process is shown on the T–s
and h–s diagram in Fig. 9.31.
1
1.5 MPa
1.5 MPa
300∞C
C.S.
40∞C
Fig. 9.30
300∞C
300∞C
1
h
1.5 MPa
a
4 kP
7.38
40∞C
2
2
x2
s
s
(a)
(b)
Fig. 9.31
From Table A.1(a), at 40°C
psat = 7.384 kPa,
sf = 0.5725,
hf = 167.57,
hfg = 2406.7 kJ/kg
and
and
sfg = 7.6845 kJ/kg K
At p = 1.5 MPa, t = 300°C, from the tabulated properties of superheated steam
(Table A.2)
s1 = 6.9189 kJ/kg K
h1 = 3037.6 kJ/kg
299
Properties of Pure Substances
Since
s1 = s2
6.9189 = sf + x2 sfg40°C
= 0.5725 + x2 ¥ 7.6845
x2 =
6.3464
= 0.826 or 82.6%
7.6845
h2 = hf40°C + x2 hfg40°C
\
= 167.57 + 0.826 ¥ 2406.7 = 2152.57 kJ/kg
W = h1 – h2 = 3037.6 – 2152.57 = 885.03 kJ/kg
\
Example 9.8 Steam at 0.8 MPa, 250°C and flowing at the rate of 1 kg/s passes
into a pipe carrying wet steam at 0.8 MPa, 0.95 dry. After adiabatic mixing the
flow rate is 2.3 kg/s. Determine the condition of steam after mixing.
The mixture is now expanded in a frictionless nozzle isentropically to a pressure
of 0.4 MPa. Determine the velocity of the steam leaving the nozzle. Neglect the
velocity of steam in the pipeline.
Solution
Figure 9.32 gives the flow diagram.
w2 = w3 – w1 = 2.3 – 1.0 = 1.3 kg/s
p1 = 0.8 MPa
t1 = 250 ∞C
m1 = 1 kg/s
V4
m3 = 2.3 kg/s
p2 = 0.8 MPa
x2 = 0.95
m2 = 1 kg/s
p4 = 0.4 MPa
Fig. 9.32
The energy equation for the adiabatic mixing of the two streams gives
w1 h1 + w2 h2 = w3 h3
At 0.8 MPa, 250°C,
At 0.8 MPa, 0.95 dry
h1 = 2950.0 kJ/kg
h2 = hf + 0.95 hfg
= 721.11 + 0.95 ¥ 2048.0 = 2666.71 kJ/kg
\ From Eq. (9.10)
1 ¥ 2950 + 1.3 ¥ 2666.71 = 2.3 ¥ h3
\
Since
h3 = 2790 kJ/kg
(hg)0.8 MPa = 2769.1 kJ/kg
(9.10)
300
Engineering Thermodynamics
and h3 > hg, the state must be in the superheated region. From the steam tables,
when p = 0.8 MPa, t = 200°C
h = 2839.3 kJ/kg
When
p = 0.8 MPa, tsat = 170.43°C
hg = 2769.1 kJ/kg
By linear interpolation
t3 = 179°C
\ Degree of superheat = 179 – 170.43 = 8.57°C
\ Condition of steam after mixing = 0.8 MPa, 179°C
The energy equation for the nozzle gives
V42
2
since V3 = velocity of steam in the pipeline = 0
Steam expands isentropically in the nozzle to 0.4 MPa. By interpolation,
h3 = h4 +
s3 = 6.7087 kJ/kg K = s4
\
6.7087 = 1.7766 + x4 ¥ 5.1193
x4 = 0.964
\
h4 = 604.74 + 0.964 ¥ 2133.8 = 2660 kJ/kg
2
V4 ¥ 10–3 = 2 (h3 – h4) = 2 ¥ 130 = 260
V4 =
26 ¥ 100 = 509.9 m/s
The processes are shown on the h–s and T–s diagrams in Fig. 9.33.
Fig. 9.33
Example 9.9 Steam flows in a pipeline at 1.5 MPa. After expanding to 0.1 MPa
in a throttling calorimeter, the temperature is found to be 120°C. Find the quality
of steam in the pipeline. What is the maximum moisture at 1.5 MPa that can be
determined with this set-up if at least 5°C of superheat is required after throttling
for accurate readings?
Properties of Pure Substances
At state 2 (Fig. 9.34), when p = 0.1 MPa, t = 120°C by interpolation
h2 = 2716.2 kJ/kg, and at p = 1.5 MPa
hf = 844.89 and hfg = 1947.3 kJ/kg
h
Solution
301
1
2
4
a
MP
1.5
MPa
0.1
3
120∞C
104.63∞C
x4 x1
s
Fig. 9.34
Now,
or
h1 = h2
hf1.5MPa + x1hfg1.5MPa = h2
844.89 + x1 ¥ 1947.3 = 2716.2
1871.3
= 0.963
1947.3
p = 0.1 MPa and t = 99.63 + 5 = 104.63°C
h3 = 2685.5 kJ/kg
h3 = h4
2685.5 = 844.89 + x4 ¥ 1947.3
x1 =
When
Since
\
1840.6
= 0.948
1947.3
The maximum moisture that can be determined with this set-up is only 5.2%.
x4 =
Example 9.10 The following data were obtained with a separating and throttling calorimeter:
Pressure in pipeline
1.5 MPa
Condition after throttling
0.1 MPa, 110°C
During 5 min moisture collected in the separator
0.150 litre at 70°C
Steam condensed after throttling during 5 min
3.24 kg
Find the quality of steam in the pipeline
Solution
Now
or
\
As shown in Fig. 9.35
at 0.1 MPa, 110°C
h3 = 2696.2 kJ/kg
h3 = h2 = hf1.5 MPa + x2 hfg1.5 MPa
2696.2 = 844.89 + x2 ¥ 1947.3
x2 =
1851.31
= 0.955
1947.3
302
h
Engineering Thermodynamics
2
3
110∞C
1
Pa
M
5
1.
Pa
0.1 M
x1
x2
s
Fig. 9.35
If m1 = mass of moisture collected in separator in 5 min and m2 = mass of steam
condensed after throttling in 5 min.
x 2 m2
m1 + m2
then
x1 =
At 70°C,
vf = 0.001023 m3/kg
m1 =
150 ¥ 10 -6 m 3
= 0.1462 kg
1023 ¥ 10 -6 m 3 / kg
m2 = 3.24 kg
x1 =
\
3.1
0.955 ¥ 3.24
=
= 0.915
0.1462 + 3.24 3.3862
Example 9.11 A steam boiler initially contains 5 m3 of steam and 5 m3 of water
at 1 MPa. Steam is taken out at constant pressure until 4 m3 of water is left. What
is the heat transferred during the process?
Solution
At 1 MPa,
vf = 0.001127, and
vg = 0.1944 m3/kg
hg = 2778.1 kJ/kg
uf = 761.68, ug = 2583.6 kJ/kg
ufg = 1822 kJ/kg
The initial mass of saturated water and steam in the boiler (Fig. 9.36).
Vf
vf
+
Vg
vg
=
5
5
+
= (4.45 ¥ 103 + 25.70) kg
0.001127 0.1944
where suffix f refers to saturated water and suffix g refers to saturated vapour. Final
mass of saturated water and steam
303
Properties of Pure Substances
=
4
6
= (3.55 ¥ 103 + 30.80) kg
+
0.001127 0.1944
1 MPa
1 MPa
Vg
Vg
= 5 m3
Vf = 4 m3
Vf = 5 m3
Q
Initial
Final
(a)
(b)
Fig. 9.36
\ Mass of steam taken out of the boiler (ms)
= (4.45 ¥ 103 + 25.70) – (3.55 ¥ 103 + 30.80)
= 0.90 ¥ 103 – 5.1 = 894.9 kg
Making an energy balance, we have: Initial energy stored in saturated water and
steam + Heat transferred from the external source = Final energy stored in saturated
water and steam + Energy leaving with the steam.
or
U1 + Q = Uf + ms hg
assuming that the steam taken out is dry (x = 1)
or
4.45 ¥ 103 ¥ 761.68 + 25.70 ¥ 2583.6 + Q
= 3.55 ¥ 103 ¥ 761.68 + 30.8 ¥ 2583.6 + 894.9 ¥ 2778.1
or
Q = 894.9 ¥ 2778.1 – (0.90 ¥ 103) ¥ 761.68 + 5.1 ¥ 2583.6
= 2425000 – 685500 + 13176
= 1752,676 kJ = 1752.676 MJ
Example 9.12 A 280 mm diameter cylinder fitted with a frictionless leakproof
piston contains 0.02 kg of steam at a pressure of 0.6 MPa and a temperature of
200°C. As the piston moves slowly outwards through a distance of 305 mm, the
steam undergoes a fully-resisted expansion during which the steam pressure p and
the steam volume V are related by pV n = constant, where n is a constant. The final
pressure of the steam is 0.12 MPa. Determine (a) the value of n, (b) the work done
by the steam, and (c) the magnitude and sign of heat transfer.
Solution
Since the path of expansion (Fig. 9.37) follows the equation
pV n = C
p1 V1n = p2 V2n
304
Engineering Thermodynamics
1
p1
p2
n=
V
log 2
V1
log
Now, at 0.6 MPa, 200°C, from Tables A.2
v1 = 0.352 m3/kg
h1 = 2850.1 kJ/kg
\ Total volume, V1, at state 1
= 0.352 ¥ 0.02 = 0.00704 m3
Displaced volume
=
pV n = c
p
Taking logarithms and arranging the terms
2
V
Fig. 9.37
p 2
d ◊l
4
p
¥ (0.28)2 ¥ 0.305 = 0.0188 m3
4
\ Total volume V2 after expansion = 0.0188 + 0.00704 = 0.02584 m3
=
0.6
0.12 = log 5 = 1.24
n=
0.02584
log 3.68
log
0.00704
log
Work done by steam in the expansion process
W1 – 2 =
=
z
V2
V1
pdV =
p1V1 - p2 V2
n -1
6 ¥ 10 5 N/m 2 ¥ 0.00704 m 2 - 1.2 ¥ 10 5 N/m 2 ¥ 0.02584 m 3
1.24 - 1
Now
4224 - 3100.8
Nm
0.24
= 4680 Nm = 4.68 kJ
V2 = 0.02584 m3
\
v2 =
=
0.02584
= 1.292 m3/kg
0.02
Again
or
v2 = vf 0.12MPa + x2vfg0.12MPa
1.292 = 0.0010476 + x2 ¥ 1.4271
1.291
= 0.906
1.427
At 0.12 MPa, uf = 439.3 kJ/kg, ug = 2512.0 kJ/kg
\
x2 =
305
Properties of Pure Substances
Again
u2 = 439.3 + 0.906 (2512 – 439.3)
= 2314.3 kJ/kg
h1 = 2850.1 kJ/kg
\
u1 = h1 – p1v1 = 2850.1 –
\
0.6 ¥ 10 6 ¥ 0.00704 ¥ 10 -3
0.02
= 2850.1 – 211.2 = 2638.9 kJ/kg
By the first law
Q1 – 2 = U2 – U1 + W1 – 2
= m (u2 – u1) + W1 – 2
= 0.02 (2314.3 – 2638.5) + 4.68
= – 6.484 + 4.68 = –1.804 kJ
Example 9.13 A large insulated vessel is divided into two chambers, one containing 5 kg of dry saturated steam at 0.2 MPa and the other 10 kg of steam, 0.8
quality at 0.5 MPa. If the partition between the chambers is removed and the steam
is mixed thoroughly and allowed to settle, find the final pressure, steam quality,
and entropy change in the process.
Solution The vessel is divided into chambers, as shown in Fig. 9.38
At 0.2 MPa,
vg = v1 = 0.8857 m3/kg
\
V1 = m1v1 = 5 ¥ 0.8857 = 4.4285 m3
Partition
At 0.5 MPa,
v2 = vf + x2vfg
= 0.001093 + 0.8 ¥ 0.3749
= 0.30101 m3/kg
\
V2 = m2 v2 = 10 ¥ 0.30101
= 3.0101 m3
\ Total volume,
m1 = 5 kg
x1 = 1.0
p1 = 0.2 MPa
Vm = V1 + V2
= 7.4386 m3 (of mixture)
Total mass of mixture, mm = m1 + m2
= 5 + 10 = 15 kg
\ Specific volume of mixture
vm =
Vm
7.4386
=
= 0.496 m3/kg
15
mm
By energy balance
m1 u1 + m2 u2 = m3 u3
At 0.2 MPa,
hg = h1 = 2706.7 kJ/kg
u1 = h1 – p1v1 @ 2706.7 kJ/kg
m2 = 10 kg
x2 = 0.8
p2 = 0.5 MPa
Fig. 9.38
306
Engineering Thermodynamics
At 0.5 MPa,
\
Now for the mixture
h2 = hf + x2 hfg
= 640.23 + 0.8 ¥ 2108.5 = 2327.03 kJ/kg
u2 = h2 – p2 v2 @ h2 = 2327.03 kJ/kg
5 ¥ 2706.7 + 10 ¥ 2327.03
h3 = hm =
15
= 2453.6 kJ/kg ~
- u3
h3 = 2453.6 kJ/kg = u3
v3 = 0.496 m3/kg
From the Mollier diagram, with the given values of h and v, point 3 after mixing
is fixed (Fig. 9.39)
x3 = 0.870 s3 = 6.29 kJ/kg K
p3 = 3.5 bar Ans. s4 = sg0.2MPa = 7.1271 kJ/kg K
s2 = sf 0.5MPa + 0.8sfg 0.5MPa
= 1.8607 + 0.8 ¥ 4.9606 = 5.8292 kJ/kg K
V3 = 0.496 m3/kg
p3 = 3.5 bar
h
0.5 MPa
0.2 MPa
2
3
1
x2
x3
=0
.8
s2
s3
=0
.87
s1
s
Fig. 9.39
\ Entropy change during the process
= m3 s3 – (m1 s1 + m2 s2)
= 15 ¥ 6.298 – (5 ¥ 7.1271 + 10 ¥ 5.8292) = 0.43 kJ/K
Example 9.14 Steam generated at a pressure of 6 MPa and a temperature of
400°C is supplied to a turbine via a throttle valve which reduces the pressure to
5 MPa. Expansion in the turbine is adiabatic to a pressure of 0.2 MPa, the
isentropic efficiency (actual enthalpy drop/isentropic enthalpy drop) being 82%.
The surroundings are at 0.1 MPa, 20°C. Determine the availability of steam
before and after the throttle valve and at the turbine exhaust, and calculate the
specific work output from the turbine. The K.E. and P.E. changes are negligible.
Solution
Steady flow availability y is given by
1 2
V + g (Z – Z0 )
2
where subscript 0 refers to the surroundings. Since the K.E. and P.E. changes are
negligible
Y = (h – h0) – T0 (s – s0) +
307
Properties of Pure Substances
Y1 = Availability of steam before throttling
= (h1 – h0) – T0(s1 – s0)
At 6 MPa, 400°C (Fig. 9.40)
h1 = 3177.2 kJ/kg
s1 = 6.5408 kJ/kg K
6 MPa, 400∞C
6 MPa
5 MPa
5 MPa
Throttle valve
h
1
400∞C
2
Steam
turbine
0.2 MPa
3
3s
0.2 MPa
s
(a)
(b)
Fig. 9.40
At 20°C
h0 = 83.96 kJ/kg
s0 = 0.2966 kJ/kg K
\
Y1 = (3177.2 – 83.96) – 293 (6.5408 – 0.2966)
= 3093.24 – 1829.54 = 1263.7 kJ/kg Ans.
Now
h1 = h2, for throttling
At h = 3177.2 kJ/kg and p = 5 MPa, from the superheated steam table
t2 = 390°C
by linear interpolation
s2 = 6.63 kJ/kg K
\
y2 = Availability of steam after throttling
= (h2 – h0) – T0 (s2 – s0 )
= (3177.2 – 83.96) – 293 (6.63 – 0.2966)
= 3093.24 – 1855.69 = 1237.55 kJ/kg Ans.
Decrease in availability due to throttling
= y1 – y2 = 1263.7 – 1237.55 = 26.15 kJ/kg
Now
s2 = s3s = 6.63 = 1.5301 + x3s (7.1271 – 1.5301)
UV
W
\
x 3s =
5.10
= 0.9112
5.5970
h3s = 504.7 + 0.9112 ¥ 2201.9 = 2511.07 kJ/kg
h2 – h 3s = 3177.2 – 2511.07 = 666.13 kJ/kg
308
Engineering Thermodynamics
h2 – h3 = his (h1 – h3s) = 0.82 ¥ 666.13 = 546.12 kJ/kg
h3 = 2631 kJ/kg = 504.7 + x2 ¥ 2201.7
2126.3
\
x3 =
= 0.966
2201.7
s3 = 1.5301 + 0.966 ¥ 5.597 = 6.9368
\
Y3 = Availability of steam at turbine exhaust
= (h3 – h0) – T0 (s3 – S0 )
= (2631 – 83.96) – 293 (6.9368 – 0.2966)
= 2547.04 – 1945.58 = 601.46 kJ/kg Ans.
Specific work output from the turbine
= h2 – h3 = 3177.2 – 2631 = 546.2 kJ/kg
The work done is less than the loss of availability of steam between states
2 and 3, because of the irreversibility accounted for by the isentropic efficiency.
\
\
Example 9.15 A steam turbine receives 600 kg/h of steam at 25 bar, 350°C. At
a certain stage of the turbine, steam at the rate of 150 kg/h is extracted at 3 bar,
200°C. The remaining steam leaves the turbine at 0.2 bar, 0.92 dry. During the
expansion process, there is heat transfer from the turbine to the surroundings at
the rate of 10 kJ/s. Evaluate per kg of steam entering the turbine (a) the availability of steam entering and leaving the turbine, (b) the maximum work, and (c) the
irreversibility. The atmosphere is at 30°C.
At 25 bar, 350°C
h1 = 3125.87 kJ/kg
s1 = 6.8481 kJ/kg K
At 30°C,
h0 = 125.79 kJ/kg
s0 = sf 30°C = 0.4369 kJ/kg K
At 3 bar, 200°C
h2 = 2865.5 kJ/kg
s2 = 7.3115 kJ/kg K
At 0.2 bar (0.92 dry)
hf = 251.4 kJ/kg
hfg = –2358.3 kJ/kg
sf = 0.8320 kJ/kg K
sg = 7.9085 kJ/kg K
\
h3 = 251.4 + 0.92 ¥ 2358.3 = 2421.04 kJ/kg
s3 = 0.8320 + 0.92 ¥ 7.0765 = 7.3424 kJ/kg K
The states of steam are shown in Fig. 9.41.
(a) Availability of steam entering the turbine
Y1 = (h1 – h0) – T0 (s1 – s0)
= (3125.87 – 125.79) – 303 (6.8481 – 0.4369)
= 3000.08 – 1942.60 = 1057.48 kJ/kg Ans.
Availability of steam leaving the turbine at state 2,
Y2 = (h2 – h0) – T0 (s2 – s0)
= (2865.5 – 125.79) – 303 (7.3115 – 0.4369)
= 2739.71 – 2083.00 = 656.71 kJ/kg Ans.
Solution
309
Properties of Pure Substances
m1 = 600 kg/h
p1 = 25 bar, t1 = 350∞C
Q = 10 kW
1 kg
W
m2 = 150 kg/h
p2 = 3 bar
t2 = 200∞C
0.75 kg
m3 = 450 kg/h
p3 = 0.2 bar
x3 = 0.92
0.25 kg
Fig. 9.41
Availability of steam leaving the turbine at state 3,
Y3 = (h3 – h0) – T0 (s3 – s0) = (2421.04 – 125.79) – 303 (7.3524 – 0.4369)
= 199.85 kJ/kg
Ans.
(b) Maximum work per kg of steam entering the turbine
m2
m
Y2 – 3 Y3 = 1057.48 – 0.25 ¥ 656.71 – 0.75 ¥ 199.85
m1
m1
= 743.41 kJ/kg
(c) Irreversibility
I = T0 (w2 s2 + w3 s3 – w1 s1) – Q
= 303 (150 ¥ 7.3115 + 450 ¥ 7.3424 – 600 ¥ 6.8481) – (–10 ¥ 3600)
= 303 (1096.73 + 3304.08 – 4108.86) + 36000
= 124,460.85 kJ/h
124.461 ¥ 103
= 124.461 MJ/h =
= 207.44 kJ/kg
600
Wrev = Y1 –
Example 9.16 Determine the exergy of (a) 3 kg of water at 1 bar and 90°C, (b)
0.2 kg of steam at 4 MPa, 500°C and (c) 0.4 kg of wet steam at 0.1 bar and 0.85
quality, (d) 3 kg of ice at 1 bar – 10°C. Assume a dead state of 1 bar and 300 K.
At the dead state of 1 bar, 300 K,
u0 = 113.1 kJ/kg, h0 = 113.2 kJ/kg K
v0 = 0.001005 m3/kg, s0 = 0.395 kJ/kg
Exergy of the system:
f = m[(u + p0 v – T0 s) – (u0 + p0v0 – T0s0)]
Now,
u0 + p0v0 – T0 s0 = h0 – T0 s0
= 113.2 – 300 ¥ 0.395 = – 5.3 kJ/kg
(a) For water at 1 bar, 90°C
u = 376.9 kJ/kg, h = 377 kJ/kg, v = 0.001035 m3/kg
s = 1.193 kJ/kg K
Solution
310
Engineering Thermodynamics
Since
p = p0,
u + p0v – T0 s = u + pv – T0 s = h – T0 s
= 377 – 300 ¥ 1.193 = 19.1 kJ/kg
Hence,
f = 3[19.1 – (– 5.3)] = 3 ¥ 24.4 = 73.2 kJ
(b) At p = 4 MPa, t = 500°C
u = 3099.8, h = 3446.3 kJ/kg, v = 0.08637 m3/kg
s = 7.090 kJ/kg K
u + p0 v – T0 s = 30998 100 ¥ 0.08637 – 300 ¥ 7.090 = 981.4 kJ/kg
f = 0.2 [981.4 – (– 5.3)] = 197.34 kJ
(c) At 0.1 bar, 0.85 quality,
u = 192 + 0.85 ¥ 2245 = 2100.25 kJ/kg
h = 192 + 0.85 ¥ 2392 = 2225.2 kJ/kg
s = 0.649 + 0.85 ¥ 7.499 = 7.023 kJ/kg K
v = 0.001010 + 0.85 ¥ 14.67 = 12.47 m3/kg
u + p0v – T0 s = 2100.25 + 100 ¥ 12.47 – 300 ¥ 7.023 = 1240.4 kJ/kg
f = 0.4 [1240.4 – (– 5.3)] = 498.3 kJ
(d) Since p = p0,
f = U – U0 + p0 (V – V0) – T0(S – S0)
= H – H0 – V( p – p0) – T0 (S – S0 )
= m[(h – h0) – T0 (s – s0)]
At 100 kPa, – 10°C
h = –354.1 kJ/kg and s = –1.298 kJ/kg K
f = 3 [–354.1 – 113.2 – 300 (–1.289 – 0.0395)] = 81.2 kJ
Example 9.17 A flow of hot water at 90°C is used to heat relatively cold water
at 25°C to a temperature of 50°C in a heat exchanger. The cold water flows at the
rate of 1 kg/s. When the heat exchanger is operated in the parallel mode, the exit
temperature of the hot water stream must not be less than 60°C. In the counterflow
operation, the exit temperature of hot water can be as low as 35°C. Compare the
second law efficiency and the rate of exergy destruction in the two modes of operation. Take T0 = 300 K.
Solution
Given:
th1 = 90°C, tc1 = 25°C,
m� c = 1 kg/s, T0 = 300 K
tc2 = 60°C,
311
Properties of Pure Substances
The two modes of operation of (a) parallel flow and (b) counterflow are shown
in Fig. 9.42.
mh
mc
t h1
mh
mh
mc
mc
mh
mc
th1
mh
tc2
t h2
tc2
t
mh
th2
t
mh
tc1
mc
t c1
A0
A0
(a)
(b)
Fig. 9.42
In parallel flow mode (a), th2 = 60°C. Neglecting any heat loss,
m� h ch (th1 – th2) = m� h cc(tc2 – tc1)
m� h (90 – 60) = 1(50 – 25)
m� h = 0.833 kg/s
In counterflow mode, th2 = 35°C,
m� h (90 – 35) = 1(50 – 25)
25
= 0.454 kg/s
55
Thus, the counterflow arrangement uses significantly less hot water.
Assuming that the hot water stream on exit from the heat exchanger is simply
dumped into the drain, the energy flow rate of the hot water stream at entry is considered as the exergy input rate to the process.
m� h =
� h [(h1 – h0) – T0 (s1 – s0)]
af1 = m
At 300 K or 27°C,
At 90°C,
Parallel flow:
At 60°C,
At 25°C,
At 50°C,
h0 = 113.2 kJ/kg and s0 = 0.395 kJ/kg K
h1 = 376.92 kJ/kg, s1 = 1.1925 kJ/kgK
af1 = 0.833 [(376.92 – 113.2) – 300 (1.1925 – 0.395)]
= 0.833 (263.72 – 239.25) = 20.38 kW
h2 = 251.13 kJ/kg,
h3 = 104.89 kJ/kg,
h4 = 209.33 kJ/kg,
s2 = 0.8312 kJ/kg K
s3 = 0.3674 kJ/kg K
s4 = 0.7038 kJ/kg K
312
Engineering Thermodynamics
Rate of exergy gain:
� c [(h4 – h3) – T0 (s4 – s3)]
=m
= 1 [209.33 – 104.89) – 300(0.7038 – 0.3674)]
= 104.44 – 100.92 = 3.52 kW
3.52
= 0.172 or 17.2%
20.38
Rate of exergy loss by hot water:
(h II)p =
= m� h [(h1 – h2) – T0 (s1 – s2)]
= 0.833 [(376.92 – 251.13) – 300 (1.1925 – 0.8312)]
= 0.833 (125.79 – 108.39) = 14.494 kW
Rate of irreversibility or exergy destruction:
= 14.494 – 3.52 = 10.974 kW Ans.
If the hot water stream is not dumped to the drain,
hII, P =
3.52
= 0.243
14.494
or
24.3%
Counterflow:
At 35°C,
h2 = 146.68 kJ/kg, s2 = 0.5053 kJ/kg K
� c [(h4 – h3) – T0(s4 – s3)] = 3.52 kW
Rate of exergy gain of cold water = m
(same as in parallel flow)
Rate of exergy input (if existing hot water is dumped to the surroundings)
= 0.454 (263.72 – 239.25) = 11.11 kW
3.52
= 0.3168 or 31.68%
1111
.
Rate of exergy loss of hot water:
hII, C =
= m� h [(h1 – h2) – T0(s1 – s2)]
= 0.454 [(376.92 – 146.68) – 300 (1.1925 – 0.5053)]
= 0.454 (230.24 – 206.16) = 10.94 kW
3.52
= 0.3217 or 32.17%
10.94
Rate of irreversibility or exergy destruction:
= 10.94 – 3.52 = 7.42 kW
The second law efficiency for the counterflow arrangement is significantly higher
and the rate of irreversibility is substantially lower compared to the parallel flow
arrangement.
hII, C =
Example 9.18 A small geothermal well in a remote desert area produces
50 kg/h of saturated steam vapour at 150°C. The environment temperature is 45°C.
This geothermal steam will be suitably used to produce cooling for homes at 23°C.
The steam will emerge from this system as saturated liquid at 1 atm. Estimate the
maximum cooling rate that could be provided by such a system.
313
Properties of Pure Substances
Solution
The energy balance of the control volume as shown in Fig. 9.43 gives:
Q� + wh1 = Q� 0 + wh2
The entropy balance is:
LM
N
OP LM
Q N
Q�
Q�
+ ws1
S�gen = 0 + ws2 T0
T
OP
Q
where T is the temperature maintained in the homes.
Home
T = 296 K
Q
wh1
150∞C
System
T
wh2
Geothermal
steam
2
1
100∞C
Q0
Environment
T0 = 218 K
s
(a)
(b)
Fig. 9.43
Solving for Q� ,
b
g b
g
bT / T g - 1
.
w h1 - T0 s1 - h2 - T0 s2 - T0 S gen
Q� =
0
.
By second law, Sgen > 0.
Therefore, for a given discharge state 2, the maximum Q� would be
b
g
w b1 - b2
Q� max =
T0 / T - 1
b
State-1:
g
T1 = 150°C = 423 K, saturated vapour
h1 = 2746.4 kJ/kg
s1 = 6.8387 kJ/kg K
State-2:
T2 – 100°C = 373 K, saturated liquid
h2 = 419.0 kJ/kg
s2 = 1.3071 kJ/kg K
So since
T0 = 318 K,
b1 = h1 – T0s1 = 2746.4 – 318 ¥ 6.8387 = 571.7 kJ/kg
b2 – h2 – T0s2 = 419.0 – 318 ¥ 1.3071 = 3.3 kJ/kg
a
f
50 ¥ 571.7 - 3.3
Q� max =
= 3.82 ¥ 105 kJ/h = 106 kW
a318 / 296f - 1
314
Engineering Thermodynamics
Review Questions
9.1 What is a pure substance?
9.2 What are saturation states?
9.3 What do you understand by triple point?
9.4 What is the critical state? Explain the terms critical pressure, critical temperature
and critical volume of water?
9.5 What is normal boiling point.
9.6 Draw the phase equilibrium diagram on p-v coordinates for a substance which
shrinks in volume on melting and then for a substance which expands in volume
on melting. Indicate thereon the relevant constant property lines.
9.7 Draw the phase equilibrium diagram for a pure substance on p-T coordinates.
Why does the fusion line for water have negative slope?
9.8 Draw the phase equilibrium diagram for a pure substance on T-s plot with
relevant constant property lines.
9.9 Draw the phase equilibrium diagram for a pure substance on h-s plot with
relevant constant property lines.
9.10 Why do the isobars on Mollier diagram diverge from one another?
9.11 Why do isotherms on Mollier diagram become horizontal in the superheated
region at low pressures?
9.12 What do you understand by the degree of superheat and the degree of subcooling?
9.13 What is quality of steam? What are the different methods of measurement of
quality?
9.14 Why cannot a throttling calorimeter measure the quality if the steam is very wet?
How is the quality measured then?
9.15 What is the principle of operation of an electrical calorimeter?
Problems
9.1 Complete the following table of properties for 1 kg of water (liquid, vapour or
mixture)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
p
(bar)
t
(°C)
v
(m3/kg)
x
(%)
Superheat (°C)
h
(kJ/kg)
s
(kJ/kg K)
—
—
—
1
10
5
4
—
20
15
35
—
212.42
—
320
—
—
500
—
—
25.22
0.001044
—
—
—
0.4646
0.4400
—
—
—
—
—
90
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
50
—
—
419.04
—
—
—
—
—
3445.3
—
—
—
—
—
6.104
—
—
—
—
—
7.2690
Properties of Pure Substances
315
9.2 (a) A rigid vessel of volume 0.86 m3 contains 1 kg of steam at a pressure of 2 bar.
Evaluate the specific volume, temperature, dryness fraction, internal energy,
enthalpy, and entropy of steam.
(b) The steam is heated to raise its temperature to 150°C. Show the process on a
sketch of the p–v diagram, and evaluate the pressure, increase in enthalpy,
increase in internal energy, increase in entropy of steam, and the heat transfer.
Evaluate also the pressure at which the steam becomes dry saturated.
Ans. (a) 0.86 m3/kg, 120.23°C, 0.97, 2468.54 k/kg,
2640.54 kJ/kg, 6.9592 kJ/kg K
(b) 2.3 bar, 126 kJ/kg, 106.6 kJ/kg, 0.2598 kJ/kg K, 106.6 kJ/K
9.3 Ten kg of water at 45°C is heated at a constant pressure of 10 bar until it becomes
superheated vapour at 300°C. Find the change in volume, enthalpy, internal
energy and entropy.
Ans. 2.569 m3, 28627.5 kJ, 26047.6 kJ, 64.842 kJ/K
9.4 Water at 40°C is continuously sprayed into a pipeline carrying 5 tonnes
of steam at 5 bar, 300°C per hour. At a section downstream where the pressure is
3 bar, the quality is to be 95%. Find the rate of water spray in kg/h.
Ans. 912.67 kg/h
9.5 A rigid vessel contains 1 kg of a mixture of saturated water and saturated steam
at a pressure of 0.15 MPa. When the mixture is heated, the state passes through
the critical point. Determine (a) the volume of the vessel, (b) the mass of liquid
and of vapour in the vessel initially, (c) the temperature of the mixture when the
pressure has risen to 3 MPa, and (d) the heat transfer required to produce the
final state (c).
Ans. (a) 0.003155 m3, (b) 0.9982 kg, 0.0018 kg,
(c) 233.9°C, (d) 581.46 kJ/kg
9.6 A rigid closed tank of volume 3 m3 contains 5 kg of wet steam at a pressure of
200 kPa. The tank is heated until the steam becomes dry saturated. Determine the
final pressure and the heat transfer to the tank.
Ans. 304 kPa, 3346 kJ
9.7 Steam flows through a small turbine at the rate of 5000 kg/h entering at 15 bar,
300°C and leaving at 0.1 bar with 4% moisture. The steam enters at 80 m/s at a
point 2 m above the discharge and leaves at 40 m/s. Compute the shaft power
assuming that the device is adiabatic but considering kinetic and potential
energy changes. How much error would be made if these terms were neglected?
Calculate the diameters of the inlet and discharge tubes.
Ans. 765.6 kW, 0.44%, 6.11 cm, 78.9 cm
9.8 A sample of steam from a boiler drum at 3 MPa is put through a throttling calorimeter in which the pressure and temperature are found to be 0.1 MPa, 120°C.
Find the quality of the sample taken from the boiler.
Ans. 0.951
9.9 It is desired to measure the quality of wet steam at 0.5 MPa. The quality of steam
is expected to be not more than 0.9.
(a) Explain why a throttling calorimeter to atmospheric pressure will not serve
the purpose.
(b) Will the use of a separating calorimeter, ahead of the throttling calorimeter, serve the purpose, if at best 5 C degree of superheat is desirable at the
end of throttling? What is the minimum dryness fraction required at the exit
of the separating calorimeter to satisfy this condition?
Ans. 0.97
316
Engineering Thermodynamics
9.10 The following observations were recorded in an experiment with a combined
separating and throttling calorimeter:
Pressure in the steam main—15 bar
Mass of water drained from the separator—0.55 kg
Mass of steam condensed after passing through the throttle valve —4.20 kg
Pressure and temperature after throttling—1 bar, 120°C
Evaluate the dryness fraction of the steam in the main, and state with reasons,
whether the throttling calorimeter alone could have been used for this test.
Ans. 0.85
9.11 Steam from an engine exhaust at 1.25 bar flows steadily through an electric calorimeter and comes out at 1 bar, 130°C. The calorimeter has two 1 kW heaters and
the flow is measured to be 3.4 kg in 5 min. Find the quality in the engine exhaust.
For the same mass flow and pressures, what is the maximum moisture that can be
determined if the outlet temperature is at least 105°C?
Ans. 0.944, 0.921
9.12 Steam expands isentropically in a nozzle from 1 MPa, 250°C to 10 kPa. The
steam flow rate is 1 kg/s. Find the velocity of steam at the exit from the nozzle,
and the exit area of the nozzle. Neglect the velocity of steam at the inlet to the
nozzle.
The exhaust steam from the nozzle flows into a condenser and flows out as saturated water. The cooling water enters the condenser at 25°C and leaves at 35°C.
Determine the mass flow rate of cooling water.
Ans. 1224 m/s, 0.0101 m2, 47.81 kg/s
9.13 A reversible polytropic process, begins with steam at p1 = 10 bar, t1 = 200°C, and
ends with p2 = 1 bar. The exponent n has the value 1.15. Find the final specific
volume, the final temperature, and the heat transferred per kg of fluid.
9.14 Two streams of steam, one at 2 MPa, 300°C and the other at 2 MPa, 400°C,
mix in a steady flow adiabatic process. The rates of flow of the two streams are 3
kg/min and 2 kg/min respectively. Evaluate the final temperature of the emerging stream, if there is no pressure drop due to the mixing process. What would be
the rate of increase in the entropy of the universe? This stream with a negligible
velocity now expands adiabatically in a nozzle to a pressure of 1 kPa. Determine
the exit velocity of the stream and the exit area of the nozzle.
Ans. 340°C, 0.042 kJ/K min, 1530 m/s, 53.77 cm2
9.15 Boiler steam at 8 bar, 250°C, reaches the engine control valve through a pipeline
at 7 bar, 200°C. It is throttled to 5 bar before expanding in the engine to 0.1 bar,
0.9 dry. Determine per kg of steam (a) the heat loss in the pipeline, (b) the temperature drop in passing through the throttle valve, (c) the work output of the
engine, (d) the entropy change due to throttling and (e) the entropy change in
passing through the engine.
Ans. (a) 105.3 kJ/kg, (b) 5°C, (c) 499.35 kJ/kg,
(d) 0.1433 kJ/kg K, (e) 0.3657 kJ/kg K
9.16 Tank A (Fig. 9.44) has a volume of 0.1 m3 and contains steam at 200°C, 10%
liquid and 90% vapour by volume, while tank B is evacuated. The valve is then
opened, and the tanks eventually come to the same pressure, which is found to be
4 bar. During this process, heat is transferred such that the steam remains at
200°C. What is the volume of tank B?
Properties of Pure Substances
A
317
B
Fig. 9.44
Ans. 4.89 m3
9.17 Calculate the amount of heat which enters or leaves 1 kg of steam initially at
0.5 MPa and 250°C, when it undergoes the following processes:
(a) It is confined by a piston in a cylinder and is compressed to 1 MPa and
300°C as the piston does 200 kJ of work on the steam.
(b) It passes in steady flow through a device and leaves at 1 MPa and 300°C
while, per kg of steam flowing through it, a shaft puts in 200 kJ of work.
Changes in K.E. and P.E. are negligible.
(c) It flows into an evacuated rigid container from a large source which is maintained at the initial condition of the steam. Then 200 kJ of shaft work is
transferred to the steam, so that its final condition is 1 MPa and 300°C.
Ans. (a) – 130 kJ (b) – 109 kJ, and (c) – 367 kJ
9.18 A sample of wet steam from a steam main flows steadily through a partially open
valve into a pipeline in which is fitted an electric coil. The valve and the pipeline
are well insulated. The steam mass flow rate is 0.008 kg/s while the coil takes
3.91 amperes at 230 volts. The main pressure is 4 bar, and the pressure and temperature of the steam downstream of the coil are 2 bar and 160°C respectively.
Steam velocities may be assumed to be negligible.
(a) Evaluate the quality of steam in the main.
(b) State, with reasons, whether an insulted throttling calorimeter could be used
for this test.
Ans. (a) 0.97, (b) not suitable
9.19 Two insulated tanks, A and B, are connected by a valve. Tank A has a volume of
0.70 m3 and contains steam at 1.5 bar, 200°C. Tank B has a volume of 0.35 m3
and contains steam at 6 bar with a quality of 90%. The valve is then opened, and
the two tanks come to a uniform state. If there is no heat transfer during the
process, what is the final pressure? Compute the entropy change of the universe.
Ans. 322.6 KPa, 0.1985 kJ/K
9.20 A spherical aluminium vessel has an inside diameter of 0.3 m and a 0.62 cm
thick wall. The vessel contains water at 25°C with a quality of 1%. The vessel is
then heated until the water inside is saturated vapour. Considering the vessel and
water together as a system, calculate the heat transfer during this process. The
density of aluminium is 2.7 g/cm3 and its specific heat is 0.896 kJ/kg K.
Ans. 2682.82 kJ
9.21 Steam at 10 bar, 250°C flowing with negligible velocity at the rate of 3 kg/min
mixes adiabatically with steam at 10 bar, 0.75 quality, flowing also with negligible velocity at the rate of 5 kg/min. The combined stream of steam is throttled
to 5 bar and then expanded isentropically in a nozzle to 2 bar. Determine (a) the
state of steam after mixing, (b) the state of steam after throttling, (c) the increase
in entropy due to throttling, (d) the velocity of steam at the exit from the nozzle,
318
Engineering Thermodynamics
and (e) the exit area of the nozzle. Neglect the K.E. of steam at the inlet to the
nozzle.
Ans. (a) 10 bar, 0.975 dry, (b) 5 bar, 0.894 dry,
(c) 0.2669 kJ/kg K, (d) 540 m/s, (e) 1.864 cm2
9.22 Steam of 65 bar, 400°C leaves the boiler to enter a steam turbine fitted with a
throttle governor. At a reduced load, as the governor takes action, the pressure of
steam is reduced to 59 bar by throttling before it is admitted to the turbine. Evaluate the availabilities of steam before and after the throttling process and the irreversibility due to it.
Ans. I = 21 kJ/kg
9.23 A mass of wet steam at temperature 165°C is expanded at constant quality 0.8 to
pressure 3 bar. It is then heated at constant pressure to a degree of superheat of
66.5°C. Find the enthalpy and entropy changes during expansion and during heating. Draw the T–s and h–s diagrams.
Ans. – 59 kJ/kg, 0.163 kJ/kgK during expansion and 676 kJ/kg,
1.588 kJ/kg K during heating
9.24 Steam enters a turbine at a pressure of 100 bar and a temperature of 400°C. At
the exit of the turbine the pressure is 1 bar and the entropy is 0.6 J/g K greater
than that at inlet. The process is adiabatic and changes in KE and PE may be
neglected. Find the work done by the steam in J/g. What is the mass flow rate of
steam required to produce a power output of 1 kW?
Ans. 625 J/g, 1.6 kg/s
9.25 One kg of steam in a closed system undergoes a thermodynamic cycle composed
the following reversible processes: (1–2) The steam initially at 10 bar, 40% quality is heated at constant volume until the pressure rises to 35 bar; (2–3). It is then
expanded isothermally to 10 bar; (3–1). It is finally cooled at constant pressure
back to its initial state. Sketch the cycle on T–s coordinates, and calculate the
work done, the heat transferred, and the change of entropy for each of the three
processes. What is the thermal efficiency of the cycle?
Ans. 0; 1364 kJ; 2.781 kJ/K, 367.5 kJ; 404.6 kJ; 0.639 kJ/K;
– 209.1 kJ; – 1611 kJ; – 3.419 kJ/K 8.93%
9.26 Determine the exergy per unit mass for the steady flow of each of the following:
(a) steam at 1.5 MPa, 500°C
(b) air at 1.5 MPa, 500°C
(c) water at 4MPa, 300K
(d) air at 4 MPa, 300 K
(e) air at 1.5 MPa, 300 K
Ans. (a) 1220 kJ/kg, (b) 424 kJ/kg, (c) 3.85 kJ/kg;
(d) 316 kJ/kg, (e) 232 kJ/kg
9.27 A liquid (cp = 6 kJ/kg K) is heated at an approximately constant pressure from
298 K to 90°C by passing it through tubes immersed in a furnace. The mass flow
rate is 0.2 kg/s. Determine (a) the heating load in kW. (b) the exergy
production rate in kW corresponding to the temperature rise of the fluid.
Ans. (a) 78 kW, (b) 7.44 kW
9.28 A flow of hot water at 80°C is used to heat cold water from 20°C to 45°C in a
heat exchanger. The cold water flows at the rate of 2 kg/s. When operated in
parallel mode, the exit temperature of hot water stream cannot be less than 55°C,
while in the counterflow mode, it can be as low as 30°C. Assuming the surroundings are at 300 K, compare the second law efficiencies for the two modes
of operation.
Properties of Pure Substances
319
9.29 Water at 90°C is flowing in a pipe. The pressure of the water is 3 bar, the mass
flow rate is 10 kg/s, the velocity is 0.5 m/s and the elevation of the pipe is 200 m
above the exit plane of the pipeline (ground level). Compute (a) the thermal
exergy flux, (b) the pressure exergy flux, (c) the exergy flux from KE, (d) the
exergy flux from PE, (e) total exergy flux of the stream.
Ans. (a) 260 kW, (b) 2.07 kW, (c) 1.25 ¥ 103 kW,
(d) 19.6 kW, (e) 282 kW
9.30 A cylinder fitted with a piston contains 2 kg steam at 500 kPa, 400°C. Find the
entropy change and the work done when the steam expands to a final pressure of
200 kPa in each of the following ways: (a) adiabatically and reversibly,
(b) adiabatically and irreversibly to an equilibrium temperature of 300°C.
Ans. (a) 0, 386.7 kJ, (b) 0.1976 kJ/K, 309.4 kJ
9.31 Steam expands isentropically in a nozzle from 1 MPa, 250°C to 10 kPa. The
steam flow rate is 1 kg/s. Neglecting the KE of steam at inlet to the nozzle, find
the velocity of steam at exit from the nozzle and the exit area of the nozzle.
Ans. 1223 m/s, 100 cm2
9.32 Hot helium gas at 800°C is supplied to a steam generator and is cooled to 450°C
while serving as a heat source for the generation of steam. Water enters
the steam generator at 200 bar, 250°C and leaves as superheated steam at
200 bar, 500°C. The temperature of the surroundings is 27°C. For 1 kg helium,
determine (a) the maximum work that could be produced by the heat removed
from helium, (b) the mass of steam generated per kg of helium, (c) the actual
work done in the steam cycle per kg of helium, (d) the net change for entropy
of the universe, and (e) the irreversibility. Take the average cp for helium as
5.1926 kJ/kg K and the properties of water in inlet to the steam generator as
those of saturated water at 250°C.
Ans. (a) 1202.4 kJ/kg He, (b) 0.844 kg H2O/kg He (c) 969.9 kJ/kg He,
(d) 0.775 kJ/(kg He-K), (e) 232.5 kJ/kg He
10
Properties of Gases and
Gas Mixtures
10.1
AVOGADRO’S LAW
A mole of a substance has a mass numerically equal to the molecular weight of the
substance.
One g mol of oxygen has a mass of 32 gm, 1 kg mol of oxygen has a mass of
32 kg, 1 kg mol of nitrogen has a mass of 28 kg, and so on.
Avogadro’s law states that the volume of a g mol of all gases at the pressure of
760 mm Hg and temperature of 0°C is the same, and is equal to 22.4 litres. Therefore, 1 g mol of a gas has a volume of 22.4 ¥ 103 cm3 and 1 kg mol of a gas has a
volume of 22.4 m3 at normal temperature and pressure (N.T.P.).
For a certain gas, if m is its mass in kg, and m its molecular weight, then the
number of kg moles of the gas, n, would be given by
n=
m
m kg
=
kg moles
kg
m
m
kg mol
The molar volume, v , is given by
V 3
m /kg mol
n
where V is the total volume of the gas in m3.
v =
10.2
EQUATION OF STATE OF A GAS
The functional relationship among the properties, pressure p, molar or specific volume v, and temperature T, is known as an equation of state, which may be expressed
in the form,
f ( p, v, T ) = 0
If two of these properties of a gas are known, the third can be evaluated from the
equation of state.
It was discussed in Chapter 2 that gas is the best-behaved thermometric substance because of the fact that the ratio of pressure p of a gas at any temperature to
Properties of Gases and Gas Mixtures
321
pressure pt of the same gas at the triple point, as both p and pt approach zero,
approaches a value independent of the nature of the gas. The ideal gas temperature
T of the system at whose temperature the gas exerts pressure p (Article 2.5) was
defined as
T = 273.16 lim
p
pt Æ0 p
t
T = 273.16 lim
V
pp Æ0 V
t
(Const. V)
(Const. p)
The relation between pv and p of a gas may be expressed by means of a power
series of the form
pv = A(1 + B¢p + C ¢p2 + ...)
(10.1)
where A, B ¢, C ¢, etc. depend on the temperature and nature of the gas.
A fundamental property of gases is that lim (pv) is independent of the nature of
pÆ0
the gas and depends only on T. This is shown in Fig. 10.1, where the product pv is
plotted against p for four different gases in the bulb (nitrogen, air, hydrogen, and
oxygen) at the boiling point of sulphur, at steam point and at the triple point of
water. In each case, it is seen that as p Æ 0, pv approaches the same value for all
gases at the same temperature. From Eq. (10.1)
lim pv = A
pÆ 0
Therefore, the constant A is a function of temperature only and independent of the
nature of the gas.
lim
A
p
pV
lim pv
(Const. V) = lim
=
=
pt
pt V
lim ( pv ) t
At
lim
A
V
pV
lim pv
(Const. p) = lim
=
=
Vt
pVt
lim( pv) t
At
The ideal gas temperature T, is thus
T = 273.16
\
lim ( pv) =
lim ( pv )
lim ( pv )t
LM lim ( pv ) OP T
N 273.16 Q
t
The term within bracket is called the universal gas constant and is denoted by R .
Thus,
R =
lim( pv) t
27316
.
(10.2)
322
Engineering Thermodynamics
pv, litre atm/gmol
lim (pv) = 58.9 litre atm/gmol
p
o
sulphur
60
N2
59.5
Air
H2, O2
59
0
10
20
30
p, atm
40
pv, litre atm/gmol
(a)
lim (pv) = 30.62 litre atm/gmol
p
o
steam
H
2
30
N2
Air
O2
30.5
30
0
10
20
30
p, atm
40
pv, litre atm/gmol
(b)
lim (pv) t = 22.42 litre atm/gmol
p
o
H2
23
22
N2
Air
21
O2
20
0
10
20
30
p, atm
40
(c)
Fig. 10.1
For Any, gas lim (pV)T is Independent of the Nature of the Gas
pÆ0
and Depends Only on T
The value obtained for lim (pv)t is 22.4
pÆ0
\
R =
litre – atm
g mol
22.4
litre – atm
= 0.083
273.16
g mol K
The equation of state of a gas is thus
lim pv = RT
p→0
where v is the molar volume.
(10.3)
Properties of Gases and Gas Mixtures
323
10.3 IDEAL GAS
A hypothetical gas which obeys the law pv = RT at all pressures and temperatures
is called an ideal gas.
Real gases do not conform to this equation of state with complete accuracy. As p
Æ 0, or T Æ •, the real gas approaches the ideal gas behaviour. In the equation pv
= RT , as T Æ 0, i.e. t Æ – 273.15°C, if v remains constant, p Æ 0, or if p remains
constant v Æ 0. Since negative volume or negative pressure is inconceivable, the
lowest possible temperature is 0 K or – 273.15°C. T is, therefore, known as the
absolute temperature.
There is no permanent or perfect gas. At atmospheric condition only, these gases
exist in the gaseous state. They are subject to liquefication or solidification, as the
temperature and pressure are sufficiently lowered.
From Avogadro’s law, when p = 760 mm Hg = 1.013 ¥ 105 N/m 2, T = 273.15 K,
and v = 22.4 m3/kg mol
R =
1.013 ¥ 10 5 ¥ 22.4
273.15
= 8314.3 Nm/kg mol K
= 8.3143 kJ/kg mol K
Since v = V/n, where V is the total volume and n the number of moles of the gas,
the equation of state for an ideal gas may be written as
pV = nR T
Also
n=
(10.4)
m
m
where m is the molecular weight
R
◊T
m
\
pV = m ◊
or
pV = mRT
where R = characteristic gas constant =
(10.5)
R
m
For oxygen, e.g
Ro2 =
8.3143
= 0.262 kJ/kg K
32
Rair =
8.3143
= 0.287 kJ/kg K
28.96
For air,
(10.6)
324
Engineering Thermodynamics
There are 6.023 ¥ 1023 molecules in a g mol of a substance.
This is known as Avogadro’s number (A).
\
A = 6.023 ¥ 1026 molecules/kg mol
In n kg moles of gas, the total number of molecules, N, are
N=nA
or
n = N/A
R
T
A
= NKT
where K = Boltzmann constant
pV = N
(10.7)
8314.3
R
=
= 1.38 ¥ 10–23 J/molecule K
6.023 ¥ 10 26
A
Therefore, the equation of state of an ideal gas is given by
pV = mRT
=
= nR T
= NKT
10.3.1 Specific Heats, Internal Energy, and Enthalpy of an
Ideal Gas
An ideal gas not only satisfies the equation of state pv = RT, but its specific heats
are constant also. For real gases, these vary appreciably with temperature, and little
with pressure.
The properties of a substance are related by
Tds = du + pdv
du p
+ dv
T T
The internal energy u is assumed to be a function of T and v, i.e.
u = f (T, v)
or
ds =
or
du =
FG ∂u IJ dT + FG ∂u IJ dv
H ∂T K
H ∂v K
v
(10.8)
(10.9)
T
From Eqs. (10.8) and (10.9)
ds =
1 ∂u
1
dT +
T ∂T v
T
FG IJ
H K
LMFG ∂u IJ + pOP dv
NH ∂v K Q
(10.10)
T
Again, let
s = f (T, v)
ds =
FG ∂s IJ dT + FG ∂s IJ dv
H ∂T K H ∂v K
v
T
(10.11)
325
Properties of Gases and Gas Mixtures
Comparing Eqs. (10.10) and (10.11)
FG ∂s IJ = 1 FG ∂u IJ
H ∂T K T H ∂T K
FG ∂s IJ = 1 LMFG ∂u IJ + pOP
H ∂v K T NH ∂v K Q
v
(10.12)
v
T
(10.13)
T
Differentiating Eq. (10.12) with respect to v when T is constant
1 ∂ 2u
∂2s
=
∂T∂v
T ∂T∂v
(10.14)
Differentiating Eq. (10.13) with respect to T when v is constant
∂2s
1 ∂2u
1 ∂p
1 ∂u
p
- 2
- 2
+
=
∂v ∂T
T ∂v∂T T ∂T v T ∂v T T
FG IJ
H K
FG IJ
H K
(10.15)
From Eqs. (10.14) and (10.15)
1 ∂ 2u
1 ∂ 2u
1 ∂p
=
+
T ∂T ◊ ∂v T ∂v ◊ ∂T T ∂T
FG IJ - 1 FG ∂u IJ - p
H K T H ∂v K T
FG ∂u IJ + p = T FG ∂p IJ
H ∂v K
H ∂T K
or
T
2
v
2
T
(10.16)
v
For an ideal gas
pv = RT
v
FG ∂p IJ = R
H ∂T K
FG ∂p IJ = R = p
H ∂T K v T
v
(10.17)
v
From Eqs. (10.16) and (10.17)
FG ∂u IJ = 0
H ∂v K
(10.18)
T
Therefore, u does not change when v changes at constant temperature. Similarly, if
u = f (T, p), it can be shown that
F ∂u I = 0. Therefore, u does not change with p
GH ∂p JK
T
either, when T remains constant.
u does not change unless T changes.
Then
u = f (T )
only for an ideal gas. This is known as Joule’s law.
(10.19)
326
If
Engineering Thermodynamics
u = f (T, v)
du =
FG ∂u IJ dT + FG ∂u IJ dv
H ∂T K
H ∂v K
v
T
Since the last term is zero by Eq. (10.18), and by definition
cv =
FG ∂u IJ
H ∂T K
v
du = cv dT
(10.20)
The equation du = cv dT holds good for an ideal gas for any process, whereas for
any other substance it is true for a constant volume process only.
Since cv is constant for an ideal gas,
Du = cv DT
The enthalpy of any substance is given by
h = u + pv
For an ideal gas
h = u + RT
Therefore
h = f (T )
(10.21)
only for an ideal gas.
Now
dh = du + RdT
Since R is a constant
Dh = Du + RDT
= cv DT + RDT
= (cv + R) DT
Since h is a function of T only, and by definition
cp =
FG ∂h IJ
H ∂T K
(10.22)
p
dh = cp dT
(10.23)
or
Dh = cp DT
From Eqs. (10.22) and (10.23)
cp = cv + R
(10.24)
or
cp – cv = R
(10.25)
The equation dh = cp dT holds good for an ideal gas, even when pressure changes,
but for any other substance, this is true only for a constant pressure change.
The ratio of cp /cv is of importance in ideal gas computations, and is designated
by the symbol g, i.e.
Properties of Gases and Gas Mixtures
cp
cv
or
From Eq. (10.25)
327
=g
cp = g cv
(g – 1) cv = R
and
If R =
(10.26)
U|
|V
||
W
(10.27)
R
is substituted in Eq. (10.26)
m
R
g -1
kJ/(kg mol) (K)
gR
c p = mc p = (c p ) mol =
g -1
cv = mcv = (cv )mol =
and
U|
V|
|W
R
g -1
g R kJ/kg K
cp =
g -1
cv =
cv and cp are the molar or molal specific heats at constant volume and at
constant pressure respectively.
It can be shown by the classical kinetic theory of gases that the values of g are
5/3 for monatomic gases and 7/5 for diatomic gases. When the gas molecule contains more than two atoms (i.e. for polyatomic gases) the value of g may be taken
approximately as 4/3. The minimum value of g is thus 1 and the maximum is 1.67.
The value of g thus depends only on the molecular structure of the gas, i.e.
whether the gas is monatomic, diatomic or polyatomic having one, two or more
atoms in a molecule. It may be noted that cp and cv of an ideal gas depend only on
g and R, i.e. the number of atoms in a molecule and the molecular weight of the gas.
They are independent of temperature or pressure of the gas.
10.3.2 Entropy Change of an Ideal Gas
From the general property relations
Tds = du + pdv
Tds = dh – vdp
and for an ideal gas, du = cv dT, dh = cp dT, and pv = RT, the entropy change
between any two states 1 and 2 may be computed as given below
du p
ds =
+ dv
T T
= cv
dT
dv
+R
T
v
328
Engineering Thermodynamics
\
s2 – s1 = cv ln
Also
ds =
(10.28)
dh v
- dp
T T
= cp
or
T2
v
+ R ln 2
v1
T1
dT
dp
–R
p
T
s2 – s1 = cp ln
p
T2
– R ln 2
p1
T1
(10.29)
Since R = cp – cv, Eq. (10.29) may be written as
or
s2 – s1 = cp ln
T2
p
p
– cp ln 2 + cv ln 2
p1
p1
T1
s2 – s1 = cp ln
v2
p
+ cv ln 2
v1
p1
(10.30)
Any one of the three Eqs. (10.28), (10.29), and (10.30), may be used for computing the entropy change between any two states of an ideal gas.
10.3.3
Reversible Adiabatic Process
The general property relations for an ideal gas may be written as
Tds = du + pdv = cv dT + pdv
and
Tds = dh – vdp = cp dT – vdp
For a reversible adiabatic change, ds = 0
\
cv dT = – pdv
and
By division
cp dT = vdp
(10.31)
(10.32)
cp
vdp
=g=–
cv
pdv
or
dp
dv
+g
=0
p
v
or
d (ln p) + g d (ln v) = d (ln c)
where c is a constant.
\
ln p + g ln v = ln c
pvg = c
Between any two states 1 and 2
p1 vg1 = p2 vg2
or
FG IJ
H K
p2
v1
=
p1
v2
(10.33)
g
Properties of Gases and Gas Mixtures
329
For an ideal gas
pv = RT
From Eq. (10.33)
p = c ◊ v–g
\
c ◊ v–g ◊ v = RT
c ◊ v1–g = RT
Tvg–1 = constant
(10.34)
Between any two states 1 and 2, for a reversible adiabatic process in the case of an
ideal gas
T1 v1g –1 = T2 v2g –1
FG IJ
H K
T2
v1
=
v2
T1
or
g -1
(10.35)
Similarly, substituting from Eq. (10.33)
F cI
v= G J
H pK
p◊
1/g
in the Eq. pv = RT
c¢
= RT
p1/g
p1–(1/g ) ◊ c ¢ = RT
\
Tp (1–g )/g = constant
Between any two states 1 and 2
T1p1(1–g )/g = T2 p2(1–g )/g
T2
=
T1
\
FG p IJ
HpK
(10.36)
( g -1)/ g
2
(10.37)
1
Equations (10.33), (10.34), and (10.36) give the relations among p, v, and T in a
reversible adiabatic process for an ideal gas.
The internal energy change of an ideal gas for a reversible adiabatic process is
given by
Tds = du + pdv = 0
or
z du = - z pdv = -z vc dv
where
pvg = p1 v1g = p2 v2g = c
\
2
2
2
1
1
1
u2 – u1 = c
g
v 12-g - v11-g
p v g ◊ v1-g - p1v1g ◊ v11-g
= 2 2 2
g -1
g -1
330
Engineering Thermodynamics
=
p2 v2 - p1v1
g -1
=
RT1 T2
R( T2 - T1 )
-1
=
g -1
g - 1 T1
=
RT1
g -1
LMF p I
MNGH p JK
g -1/ g
2
1
FG
H
O
- 1P
PQ
IJ
K
(10.38)
The enthalpy change of an ideal gas for a reversible adiabatic process may be similarly derived.
Tds = dh – vdp = 0
2
2
2 ( c)1/ g
1
1
1
or
z z z
where
p1 vg1 = p2 vg2 = c
dh =
\
h2 – h1 =
=
vdp =
p1/g
g
c1/g [ p2(g–1) /g – p1(g – 1)/g ]
g -1
g
( p1 v1g )1/g ◊ (p1)(g – 1)/g
g -1
(g -1)/ g
LMF p I
MNGH p JK
(g -1)/ g
OP
PQ
-1
2
1
LMF p I - OP
1
MNGH p JK
PQ
O
g RT LF p I
M
- 1P
=
(10.39)
G
J
g - 1 MH p K
P
N
Q
The work done by an ideal gas in a reversible adiabatic process is given by
=
g p1v1
g -1
1
1
2
2
(g -1)/ g
1
d- Q = dU + d- W = 0
or
d- W = – dU
i.e. work is done at the expense of the internal energy.
\
W1–2 = U1 – U2 = m (u1 – u2)
=
LM F I
MN GH JK
mRT1
p
m( p1v1 - p2 v 2 )
mR ( T1 - T2 )
=
=
1- 2
g -1
g -1
g -1
p1
(g -1)/ g
OP
PQ
(10.40)
where m is the mass of gas.
In a steady flow process, where both flow work and external work are involved,
we have from S.F.E.E.,
Wx + D
=
V2
g R( T1 - T2 )
+ gD z = h1 – h2 = cp (T1 – T2) =
g -1
2
LM F I
MN GH JK
g ( p1v1 - p2 v2 )
p
g
=
p 1 v1 1 - 2
g -1
g -1
p1
g -1/ g
OP
PQ
(10.41)
Properties of Gases and Gas Mixtures
331
If K.E. and P.E. changes are neglected,
LM F I
MN GH JK
p
g
Wx =
p 1 v1 1 - 2
g -1
p1
10.3.4
g -1/ g
OP
PQ
(10.42)
Reversible Isothermal Process
When an ideal gas of mass m undergoes a reversible isothermal process from state
1 to state 2, the work done is given by
z
2
1
or
V2
d- W =
z
W1–2 =
V
dV = mRT ln
z mRT
V
V
V1
pdV
V2
V1
2
1
= mRT ln
p1
p2
(10.43)
The heat transfer involved in the process
Q1–2 = U2 – U1 + W1–2
= W1–2 = mRT ln V2/V1 = T (S2 – S1 )
10.3.5
(10.44)
Polytropic Process
An equation of the form pvn = constant, where n is a constant can be used approximately to describe many processes which occur in practice. Such a process is called
a polytropic process. It is not adiabatic, but it can be reversible. It may be noted that
g is a property of the gas, whereas n is not. The value of n depends upon the process. It is possible to find the value of n which more or less fits the experimental
results. For two statses on the process,
p1 v n1 = p2 v n2
(10.45)
n
or
FG v IJ = p
Hv K p
2
1
1
2
log p1 - log p2
n=
(10.46)
log v2 - log v1
For known values of p1, p2, v1 and v2, n can be estimated from the above relation.
Two other relations of a polytropic process, corresponding to Eqs. (10.35) and
(10.37), are
FG IJ
H K
Fp I
T
=G J
T
Hp K
T2
v
= 1
v2
T1
2
2
1
1
n -1
(10.47)
n -1/ n
(10.48)
332
Engineering Thermodynamics
(i) Entropy Change in a Polytropic Process In a reversible adiabatic process, the entropy remains constant. But in a reversible polytropic process, the entropy changes. Substituting Eqs. (10.45), (10.47) and (10.48) in Eqs. (10.28),
(10.29) and (10.30), we have three expressions for entropy change as given below:
T
v
s2 – s1 = cv ln 2 + R ln 2
T1
v1
=
T
T
R
R
ln 2 +
ln 1
g -1
T1 n - 1
T2
=
T
n -g
R ln 2
(g - 1)( n - 1)
T1
(10.49)
Relations in terms of pressure and specific volume can be similarly derived. These
are
p
n-g
s2 – s1 =
R ln 2
(10.50)
n(g - 1)
p1
s2 – s1 = -
and
v
n-g
R ln 2
g -1
v1
(10.51)
It can be noted that when n = g, the entropy change becomes zero. If p2 > p1, for
n £ g, the entropy of the gas decreases, and for n > g, the entropy of the gas increases. The increase of entropy may result from reversible heat transfer to the
system from the surroundings. Entropy decrease is also possible if the gas is cooled.
(ii) Heat and Work in a Polytropic Process
Using the first law to unit mass
of an ideal gas,
Q – W = u2 – u1
= cv (T2 – T1) =
=
R(T2 - T1 )
g -1
p2 v2 - p1v1
g -1
pv
= 1 1
g -1
LMF p I
MNGH p JK
n -1/ n
2
O p v LMF v I
- 1P =
PQ g - 1 MNGH v JK
1 1
1
n -1
1
2
OP
PQ
-1
(10.52)
For a steady flow system of unit mass of ideal gas, the S.F.E.E. Eq. (5.10), gives
Q – Wx – D
LM V + gz OP = h – h
N2 Q
2
2
1
= cp (T2 – T1) =
=
g R ( T2 - T1 )
g -1
g
(p2 v2 – p1v1)
g -1
(10.53)
Properties of Gases and Gas Mixtures
333
For a polytropic process,
n -1/ n
Q – Wx – D
LM V + gzOP = g p v LMFG p IJ - 1OP
PQ
N 2 Q g - 1 MNH p K
O
g p v LF v I
M
- 1P
=
G
J
g - 1 MH v K
PQ
N
2
1 1
2
1
n -1
1 1
1
(10.54)
2
Equations (10.52) and (10.54) can be used to determine heat and work quantities
for a closed and a steady flow system respectively.
(iii) Integral Property Relations in a Polytropic Process In a pvn = constant process,
2 p vn
p1v1
1 1
n dv =
1 v
n -1
z z
1
n -1
LM - F v I OP
MN1 GH v JK PQ
OP
pv L Fp I
M
=
1- G J
n -1M H p K
PQ
N
2
pdv =
1
2
n -1 / n
2
1 1
(10.55)
1
Similarly,
z
2
1
n -1
LM F I OP
MN GH JK PQ
OP
np v L F p I
M
=
1- G J
n -1 M H p K
PQ
N
- vdp =
np1v1
v
1- 1
n -1
v2
n -1 / n
2
1 1
(10.56)
1
The integral of Tds is obtained from the property relation
Tds = du + pdv
\
2
2
z z z
Tds =
1
2
du +
1
pdv
1
z
2
= u2 – u1 + 1 pdv
Substituting from Eqs. (10.50) and (10.53)
z
2
1
n -1/ n
LM F I OP
MN GH JK PQ
L Fv I O
g -n
=
p v M1 - G J P
(g - 1)( n - 1)
MN H v K PQ
Tds =
p
g -n
p1 v 1 1 - 2
(g - 1)( n - 1)
p1
n -1
1
1 1
2
334
Engineering Thermodynamics
=
g -n
R (T1 – T2)
(g - 1)( n - 1)
(10.57)
Since R/(g – 1) = cv, and putting DT = T2 – T1, the reversible heat transfer
g -n
2
QR = 1 Tds = cv
DT
1- n
= cn DT
(10.58)
where cn = c v (g – n)/(1 – n) is called the polytropic specific heat. For n > g there
will be positive heat transfer and gain in entropy. For n < g, heat transfer will be
negative and entropy of the gas would decrease.
Ordinarily both heat and work are involved in a polytropic process. To evaluate
the heat transfer during such a process, it is necessary to first evaluate the work via
either Ú pdv or – Ú vdp, depending on whether it is a closed or an open steady flow
system. The application of the first law will then yield the heat transfer.
The polytropic processes for various values of n are shown in Fig. 10.2 on the p–
v and T–s diagrams.
pvn = C
On differentiation,
vn dp + pnvn – 1 dv = 0
z
dp
p
= -n
dv
v
(10.59)
n=±•
n = ± • ( v = c)
n=–1
n=g
n=–2
n=–1
n = – 0.5
p
T
n = 0 (p = c)
n =1
n=0
n = 1 (T = C)
n = g (s = c)
v
s
n
Fig. 10.2 Process in which pv = Constant
The slope of the curve increases in the negative direction with increase of n. The
values of n for some familiar processes are given below
Isobaric process ( p = c), n = 0
Isothermal process (T = c), n = 1
Isentropic process (s = c), n = g
Isometric or isochoric process (v = c), n = •.
Properties of Gases and Gas Mixtures
335
10.4 EQUATIONS OF STATE
The ideal gas equation of state pv = RT can be established from the postulates of
the kinetic theory of gases developed by Clerk Maxwell, with two important assumptions that there is little or no attraction between the molecules of the gas and
that the volume occupied by the molecules themselves is negligibly small compared to the volume of the gas. When pressure is very small or temperature very
large, the intermolecular attraction and the volume of the molecules compared to
the total volume of the gas are not of much significance, and the real gas obeys very
closely the ideal gas equation. But as pressure increases, the intermolecular forces
of attraction and repulsion increase, and also the volume of the molecules becomes
appreciable compared to the total gas volume. So then the real gases deviate considerably from the ideal gas equation. van der Waals, by applying the laws of mechanics to individual molecules, introduced two correction terms in the equation of
ideal gas, and his equation is given below.
F p + a I (v – b) = RT
H vK
2
(10.60)
The coefficient a was introduced to account for the existence of mutual attraction
between the molecules. The term a/v2 is called the force of cohesion. The coefficient b was introduced to account for the volumes of the molecules, and is known as
co-volume.
Real gases conform more closely with the van der Waals equation of state than
the ideal gas equation of state, particularly at higher pressures. But it is not obeyed
by a real gas in all ranges of pressures and temperatures. Many more equations of
state were later introduced, and notable among these are the equations developed
by Berthelot, Dieterici, Beattie-Bridgeman, Kammerlingh Onnes, Hirshfelder-BirdSportz-McGee-Sutton, Wohl, Redlich-Kwong, and Martin-Hou.
Apart from the van der Waals equation, three two-constant equations of state are
those of Berthelot, Dieterici, and Redlich-Kwong, as given below:
Berthelot:
p=
RT
a
- 2
v - b Tv
(10.61)
Dieterici:
p=
RT
◊ e–a/RTv
v-b
(10.62)
Redlich-Kwong:
p=
RT
a
v - b T 1/ 2 v( v + b)
(10.63)
The constants, a and b are evaluated from the critical data, as shown for van der
Waals equation in Article 10.7. The Berthelot and Dieterici equations of state, like
the van der Waals equation, are of limited accuracy. But the Redlich-Kwong equation gives good results at high pressures and is fairly accurate for temperatures
above the critical value.
336
Engineering Thermodynamics
Another two-constant equation which is again of limited accuracy is the SahaBose equation of state as given below
RT –a/RTv
v - 2b
p= e
ln
(10.64)
2b
v
It is, however, quite accurate for densities less than about 0.8 times the critical
density.
One more widely used equation of state with good accuracy is the BeattieBridgeman Equation:
RT (1 - e )
A
p=
(v + B) – 2
(10.65)
v
v2
where
a
b
c
A = A0 1 , B = B0 1 - , e =
vT 3
v
v
There are five constants, A0, B 0, a, b, and c, which have to be determined
experimentally for each gas. The Beattie-Bridgeman equation does not give
satisfactory results in the critical point region.
All these equations mentioned above reduce to the ideal gas equation for large
volumes and temperatures and for very small pressures.
F
H
F
H
I
K
I
K
F
H
I
K
10.5 VIRIAL EXPANSIONS
The relations between pv and p in the form of power series, as given in
Eq. (10.1), may be expressed as
pv = A(1 + B ¢p + C ¢p2 + D ¢ p3 + ...)
For any gas, from Eq. (10.3)
lim pv = A = RT
pÆ0
pv
= 1 + B ¢p + C ¢p2 + D ¢p3 + ...
(10.66)
RT
An alternative expression is
pv
B C
D
= 1 + + 2 + 3 +�
(10.67)
v v
v
RT
Both expressions in Eqs. (10.66) and (10.67) are known as virial expansions or
virial equations of state, first introduced by the Dutch physicist, Kammerlingh
Onnes, B¢, C ¢, B, C, etc. are called virial coefficients. B ¢ and B are called second
virial coefficients, C ¢ and C are called third virial coefficients, and so on. For a
given gas, these coefficients are functions of temperature only.
\
The ratio pv / RT is called the compressibility factor, Z. For an ideal gas Z = 1.
The magnitude of Z for a certain gas at a particular pressure and temperature gives
an indication of the extent of deviation of the gas from the ideal gas behaviour. The
virial expansions become
Z = 1 + B ¢p + C ¢p2 + D ¢p3 + ...
(10.68)
Properties of Gases and Gas Mixtures
337
B C
D
(10.69)
+ 2 + 3 +�
v v
v
The relations between B¢, C ¢ and B, C, ... can be derived as given below
Z = 1+
and
pv
= 1 + B ¢p + C ¢p2 + D ¢p3 + ...
RT
LM RT F1 + B + C +...I OP
N v H v v KQ
LF RT I F1 + B + C ... +I OP + ...
+ C ¢ MG
MNH v JK H v v K PQ
= 1 + B¢
2
2
2
2
= 1+
B ¢RT B¢ BRT + C ¢ ( RT ) 2
+
v
v2
B ¢RTC + C ¢ ( R T ) 2 + D¢ ( RT ) 3
+ ...
v3
Comparing this equation with Eq. (10.81) and rearranging
+
B¢ =
B
C - B2
, C¢ =
,
RT
( RT )2
D¢ =
D - 3 BC + 2 B 3
, and so on
( RT ) 3
Z=
pv
= 1 + B ¢p + C ¢p2 + ...
RT
(10.70)
Therefore
= 1+
B
C - B2 2
p+
p + ...
RT
( RT )2
(10.71)
The terms B/ v , C/ v 2 etc. of the virial expansion arise on account of molecular
interactions. If no such interactions exist (at very low pressures) B = 0, C = 0,
etc. Z = 1 and pv = RT .
10.6 LAW OF CORRESPONDING STATES
For a certain gas, the compressibility factor Z is a function of p and T [Eq. (10.71)],
and so a plot can be made of lines of constant temperature on coordinates of p and
Z (Fig. 10.3). From this plot Z can be obtained for any value of p and T, and the
volume can then be obtained from the equation pv = ZRT. The advantage of using Z
instead of a direct plot of v is a smaller range of values in plotting.
338
Engineering Thermodynamics
35 K
1.5
100 K
50 K
60 K
200 K
1.0
300 K
Z
0.5
0
100
200
p (atm)
Fig. 10.3
Variation of the Compressibility factor of Hydrogen with
Pressure at Constant Temperature
For each substance, there is a compressibility factor chart. It would be very convenient if one chart could be used for all substances. the general shapes of the
vapour dome and of the constant temperature lines on the p–v plane are similar for
all substances, although the scales may be different. This physical similarity can be
exploited by using dimensionless properties called reduced properties. The reduced
pressure is the ratio of the existing pressure to the critical pressure of the substance,
and similarly for reduced temperature and reduced volume. Then
pr =
p
T
v
, Tr =
, vr =
pc
Tc
vc
where subscript r denotes the reduced property, and subscript c denotes the
property at the critical state.
At the same pressure and temperature the specific or molal volumes of different
gases are different. However, it is found from experimental data that at the same
reduced pressure and reduced temperature, the reduced volumes of different gases
are approximately the same. Therefore, for all substances
vr = f (pr, Tr )
(10.72)
Now,
vr =
v
ZRT pc
Z Tr
=
=
◊
Zc RTc p
Zc pr
vc
(10.73)
pc v c
. This is called the critical compressibility factor. Therefore from
RTc
Eqs. (10.72) and (10.73),
where Zc =
Properties of Gases and Gas Mixtures
Z = f (pr, Tr , Zc )
339
(10.74)
Experimental values of Zc for most substances fall within a narrow range
0.20–0.30. Therefore, Zc may be taken to be a constant and Eq. (10.74) reduces to
Z = f (pr, Tr )
(10.75)
When Tr is plotted as a function of reduced pressure and Z, a single plot, known
as the generalized compressibility chart, is found to be satisfactory for a great variety of substances. Although necessarily approximate, the plots are extremely useful in situations where detailed data on a particular gas are lacking but its critical
properties are available.
The relation among the reduced properties, pr, Tr , and vr, is known as the law of
corresponding states. It can be derived from the various equations of state, such as
those of van der Waals, Berthelot, and Dieterici. For a van der Waals gas,
F p + a I (v – b) = RT
H vK
2
where a, b, and R are the characteristic constants of the particular gas.
RT
a
\
p=
- 2
v-b v
or
pv3 – (pb + RT ) v2 + av – ab = 0
It is therefore a cubic in v and for
Critical state
given values of p and T has three roots
p
c
of which only one need be real. For
low temperatures, three positive real
p
roots exist for a certain range of presT = Tc
sure. As the temperature increases the
three real roots approach one another
and at the critical temperature they
Vc
v
become equal. Above this temperaFig. 10.4 Critical Properties
ture only one real root exists for all
on p-v Diagram
values of p. The critical isotherm Tc at
the critical state on the p–v plane (Fig. 10.4), where the three real roots of the van
der Waals Equation coincide, not only has a zero slope, but also its slope changes at
the critical state (point of inflection), so that the first and second derivatives of p
with respect to v at T = Tc are each equal to zero. Therefore
FG ∂p IJ
H ∂v K
FG ∂ p IJ
H ∂v K
= T = Tc
2
=
2
T = Tc
RTc
2a
+
=0
( vc - b ) 2 vc3
(10.76)
2 ◊ RTc
6a
=0
3 - 4
( vc - b)
vc
From these two equations, by rearranging and dividing, b =
Substituting the value of b in Eq. (10.76)
(10.77)
1
vc.
3
340
Engineering Thermodynamics
8a
9Tc vc
Substituting the values of b and R in Eq. (10.60)
R=
F p + a I F 2 v I = 8a ◊ T
GH v JK H 3 K 9T v
c
c
c
2
c
c c
a = 3pc v 2c
Therefore, the value of R becomes
R=
8 pc v c
3 Tc
The values of a, b, and R have thus been expressed in terms of critical properties.
Substituting these in the van der Waals equation of state
F p + 3p v I F v - 1 v I = 8 p v ◊ T
GH v JK H 3 K 3 T
F p + 3v I FG v - 1 IJ = 8 T
GH p v JK H v 3K 3 T
2
c c
2
c c
c
c
or,
2
c
2
c
c
c
Using the reduced parameters,
F p + 3 I (3v – 1) = 8T
GH v JK
\
r
r
2
r
(10.78)
r
In the reduced equation of state (10.78) the individual coefficients a, b and R for
a particular gas have disappeared. So this equation is an expression of the law of
corresponding states because it reduces the properties of all gases to one formula.
It is a ‘law’ to the extent that real gases obey van der Waals equation. Two different
substances are considered to be in ‘corresponding states’, if their pressures, volumes and temperatures are of the same fractions (or multiples) of the critical pressure, volume and temperature of the two substances. The generalized compressibility charts in terms of reduced properties is shown in Fig. 10.5(a) and (b). It is very
useful in predicting the properties of substances for which more precise data are not
available. The value of Z at the critical state of a van der Waals gas is 0.375
FG since R = 8 p v IJ . At very low pressures Z approaches unity, as a real gas ap3 T K
H
c c
c
proaches the ideal gas behaviour. Equation (10.78) can also be written in the following form
FG p v + 3 IJ (3v – 1) = 8T v
v K
H
r r
\
prvr =
r
8Tr vr
3
3vr - 1 vr
r
r r
(10.79)
Z
0
1.0
1.0
Liquid
Tr = 0.9
Critical point
Gas
Tr = 1.2
Saturated vapour
Saturated liquid
Pr
Foldback isotherm
Tr 5.0
Fig. 10.5 (a) Generalized Compressibility Chart
Tr = 1.0
Critical isotherm
Tr = 1.5
Line of Z = 1
Boyle isotherm
Tr 2.5
Tr = 20.0
Tr = 50.0
Tr = 1.2
Tr = 1.0
Tr = 0.9
10.0
Properties of Gases and Gas Mixtures
341
342
Engineering Thermodynamics
1.1
TR = 2.00
1.0
RT
Z =
pv
0.9
TR = 1.50
0.8
TR = 1.30
0.7
0.6
TR = 1.20
0.5
Legend
Isopentane
Methane
n-Heptane
Ethylene
Nitrogen
Ethane
Carbon dioxide
Propane
Water
n-Butane
TR = 1.10
0.4
TR = 1.00
0.3
Average curve based on
data on hydrocarbons
0.2
0.1
0
0.5
1.0
1.5
2.0
3.0
2.5
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
Reduced pressure PR
(b) Generalized Compressibility Chart
Fig. 10.5
Figure 10.6 shows the law of corresponding states in reduced coordinates, (prvr)
vs pr. Differentiating Eq. (10.79) with respect to pr and making it equal to zero, it is
possible to determine the minima of the isotherms as shown below.
LM
N
OP = 0
Q
∂ L 8T v
3 O L ∂v O
- P M
M
P =0
∂v N 3v - 1 v Q N ∂p Q
LM ∂v OP π 0
N ∂p Q
∂ L 8T v
3O
- P =0
M
∂v N 3v - 1 v Q
∂ 8Tr vr
3
∂pr 3vr - 1 vr
or
r r
r
Since
r
Tr
r
r
Tr
r
Tr
r
r
Tr
r
Tr
r r
r
r
\
8Tr
3
= 2
(3vr - 1) 2
vr
or
3(3vr - 1) 2
3
= 8Tr = pr + 2
2
vr
vr
LM
N
OP (3v – 1)
Q
r
343
Properties of Gases and Gas Mixtures
Simplifying
(prvr)2 – 9(prvr) + 6pr = 0
This is the equation of a parabola passing through the minima of the isotherms
(Fig. 10.6).
Tr = 2.8
Tr = TB = 2.54
Tr = 2
9.0
pr vr
Locus passing through the
minima of the isotherms
Tr = 1.7
Tr = 1.5
4.5
1.0
Tr = 1.2
Sat.
vap.
line
Tr = 1.0
Sat.
liq.
line
Tr = 0.8
Tr = 0.9
Critical
State
Liquid region
0
1.0
3.375
Vapour dome
pr
Fig. 10.6 Law of Corresponding States in Reduced Coordinates
When
pr = 0,
prvr = 0, 9
Again
pr =
9( pr v r ) - ( pr v r ) 2
6
dpr
= 9 – 2 ( pr v r ) = 0
d ( pr vr )
\
pr vr = 4.5
9 ¥ 4.5 - ( 4.5) 2
= 3.375
6
The parabola has the vertex at prvr = 4.5 and pr = 3.375, and it intersects the ordinate at 0 and 9.
Each isotherm up to that marked TB has a minimum (Fig. 10.6). The TB isotherm
has an initial horizontal portion (prvr = constant) so that Boyle’s law is obeyed
fairly accurately up to moderate pressures. Hence, the corresponding temperature
is called the Boyle temperature for that gas. The Boyle temperature TB can be determined by making
pr =
344
Engineering Thermodynamics
LM ∂( p v ) OP
N ∂p Q
r r
r
= 0 when pr = 0
Tr = TB
Above the Boyle temperature, the isotherms slope upward and show no minima.
As Tr is reduced below the critical (i.e. for Tr < 1), the gas becomes liquefied,
and during phase transition isotherms are vertical. The minima of all these isotherms lie in the liquid zone.
Van der Waals equation of state can be expressed in the virial form as given
below
a
p + 2 (v – b) = RT
v
a
b
pv +
1−
= RT
v
v
-1
a
b
\
pv + = RT 1 v
v
b b 2 b3
b
where < 1
= RT 1 + + 2 + 3 + ...
v v
v
v
F
H
FG
H
I
K
IJ FG
KH
IJ
K
F I
H K
LM
OP
F
H
N
Q
L F a I 1 + b + b +...OP
pv = RT M1 + b N H RT K v v v Q
\
2
3
2
3
I
K
(10.80)
\ The second virial coefficient B = b – a/RT, the third virial coefficient C = b2, etc.
From Eq. (10.71), on mass basis
FG
H
pv = RT 1 +
IJ
K
B
C - B2 2
p+
p + ...
RT
RT 2
To determine Boyle temperature, TB
LM ∂( pv) OP
N ∂p Q
T =C
p= 0
=0=
B
RT
\
B=0
a
a
or
TB =
, because B = b –
bR
RT
The point at which B is equal to zero gives the Boyle temperature. The second virial
coefficient is the most important. Since
LM ∂( pv) OP
N ∂p Q
= B, when B is known, the
p= 0
behaviour of the gas at moderate pressures is completely determined. The terms
which contain higher power (C/v2, D/v3, etc.) becomes significant only at very high
pressures.
10.7 OTHER EQUATIONS OF STATE
Van der Waals equation is one of the oldest equations of state introduced in 1899,
where the constants a and b are related to the critical state properties as found
earlier,
345
Properties of Gases and Gas Mixtures
1
1 RTc
27 R2 Tc2
, b = vc =
3
8 pc
64 pc
The Beattie-Bridgeman equation developed in 1928 is given in Eq. 10.65 which
has five constants. It is essentially an empirical curve fit of data, and is reasonably
accurate when values of specific volume are greater than vc.
Benedict, Webb, and Rubin extended the Beattie-Bridgeman equation of state to
cover a broader range of states as given below:
RT
c 1 ( bRT - a ) a
p=
+ BRT - A - 2
+
+ 6
v
T v2
v3
v
g
g
c
+ 3 2 1 + 2 exp - 2
v T
v
v
It has eight constants and is quite successful in predicting the p–v–T behaviour
of light hydrocarbons.
The Redlich-Kwong equation proposed in 1949 and given by Eq. 10.63 has the
constants a and b in terms of critical properties as follows:
R2 Tc2.5
RTc
a = 0.4275
, b = 0.0867
pc
pc
The values of the constants for the van der Waals, Redlich-Kwong and BenedictWebb-Rubin equations of state are given in Table 10.1, while those for the BeattieBridgeman equation of state are given in Table 10.2. Apart from these, many other
multi-constant equations of state have been proposed. With the advent of high speed
computers, equations having 50 or more constants have been developed for representing the p–v–T behaviour of different substances.
a = 3pc v2c =
F
H
I
K
I F I
K H K
F
H
Table 10.1
Constants for the van der Waals, Redlich-Kwong, and Benedict-Webb-Rubin
Equations of State
1. van der Waals and Redlich-Kwong: Constants for pressure in bars, specific
volume in m3/kmol, and temperature in K
van der Waals
b
a
Fm I
GH kmol JK
3
Substance
Air
Butane (C4H10)
Carbon dioxide (CO2)
Carbon monoxide (CO)
Methane (CH4)
Nitrogen (N2)
Oxygen (O2)
Propane (C3H8)
Refrigerant 12
Sulfur dioxide (SO2)
Water (H2O)
bar
2
1.368
13.86
3.647
1.474
2.293
1.366
1.369
9.349
10.49
6.883
5.531
Source: Calculated from critical data.
m3
kmol
0.0367
0.1162
0.0428
0.0395
0.0428
0.0386
0.317
0.0901
0.0971
0.0569
0.0305
Redlich-Kwong
a
b
2
Fm I K
GH kmol JK
3
bar
15.989
289.55
64.43
17.22
32.11
15.53
17.22
182.23
208.59
144.80
142.59
1/ 2
m3
kmol
0.02541
0.08060
0.02963
0.02737
0.02965
0.02677
0.02197
0.06242
0.06731
0.03945
0.02111
346
Engineering Thermodynamics
2. Benedict-Webb-Rubin:Constants for pressure in bars, specific volume in
m3/k mol, and temperature in K
Substance a
A
C4H 10 1.9073 10.218
CO 2 0.1386 2.7737
CO
0.0371 1.3590
CH 4 0.0501 1.8796
N2
0.0254 1.0676
b
0.039998
0.007210
0.002632
0.003380
0.002328
B
c
C
0.12436 3.206 ¥ 105
0.04991 1.512 ¥ 104
0.05454 1.054 ¥ 103
0.04260 2.579 ¥ 103
0.04074 7.381 ¥ 102
a
1.006 ¥ 106
1.404 ¥ 105
8.676 ¥ 103
2.287 ¥ 104
8.166 ¥ 101
g
1.101 ¥ 10–3 0.0340
8.47 ¥ 10–5 0.00539
1.350 ¥ 10–4 0.0060
1.244 ¥ 10–4 0.0060
1.272 ¥ 10–4 0.0053
Source: H.W. Copper, and J.C. Goldfrank, Hydrocarbon Processing, 45 (12); 141
(1967).
Table 10.2
(a) The Beattie-Bridgeman equation of state is
P=
Ru T
v2
FG 1 - c IJ bv + Bg - A , where A = A F1 - a I and B = B F1 - b I
H vK
H vK
H vT K
v
0
2
0
When P is in kPa, v is in m3/k mol, T is in K, and R u = 8.314 kPa ◊ m3/(kmol ◊ K), the
five constants in the Beattie-Bridgeman equation are as follows:
Gas
Air
Argon, Ar
Carbon dioxide, CO2
Helium, He
Hydrogen, H2
Nitrogen, N2
Oxygen, O 2
A0
a
B0
b
c
131.8441
130.7802
507.2836
2.1886
20.0117
136.2315
151.0857
0.01931
0.02328
0.07132
0.05984
– 0.00506
0.02617
0.02562
0.04611
0.03931
0.10476
0.01400
0.02096
0.05046
0.04624
– 0.001101
0.0
0.07235
0.0
– 0.04359
– 0.00691
0.004208
4.34 ¥ 10 4
5.99 ¥ 10 4
6.60 ¥ 10 5
40
504
4.20 ¥ 10 4
4.80 ¥ 10 4
Source: Gordon J. Van Wylen and Richard E. Sonntag, Fundamentals of Classical Thermodynamics, English/SI Version, 3d Edition., Wiley New York, 1986, p. 46. Table 3.3.
10.8 PROPERTIES OF MIXTURES OF GASES—DALTON’S LAW
OF PARTIAL PRESSURES
Let us imagine a homogeneous mixture of inert ideal gases at a temperature T, a
pressure p, and a volume V. Let us suppose there are n1 moles of gas A1, n2 moles of
gas A2, ... and upto n c moles of gas Ac (Fig. 10.7). Since there is no chemical reaction, the mixture is in a state of equilibrium with the equation of state
pV = (n 1 + n 2 + ... nc) RT
where
R = 8.3143 kJ/kg mol K
\
p=
n1 R T n2 R T
n RT
+
+�+ c
V
V
V
P
n1
T
n2
V
nc
Fig. 10.7 Mixture of Gases
Properties of Gases and Gas Mixtures
347
nk R T
represents the pressure that the kth gas would exert if it
V
occupied the volume V alone at temperature T. This is called the partial pressure of
the kth gas and is denoted by pk . Thus
The expression
p1 =
n1 RT
n RT
n RT
, p2 = 2
, ..., pc = c
V
V
V
and
p = p1 + p2 + ... + pc
(10.81)
This is known as Dalton’s law of partial pressures which states that the total pressure of a mixture of ideal gases is equal to the sum of the partial pressures.
Now
RT
V = (n1 + n 2 + ... + n c) ◊
p
RT
p
and the partial pressure of the Kth gas is
= SnK ◊
pk =
nk RT
V
Substituting the value of V
pK =
n K RT ◊ p
n RT
= K
◊p
S n K ◊ RT
 nK
Now
S nK = n1 + n2 + ... + nc
= Total number of moles of gas
nk
The ratio
is called the mole fraction of the kth gas, and is denoted by xK .
S nk
Thus
n
n
n
x1 = 1 , x2 = 2 ,..., xc = c
S nK
S nK
S nK
and
p1 = x1p, p2 = x 2 p, ..., pc = xc p
or
pK = xK ◊ p
(10.82)
Also
x1 + x2 + ... + xc = 1
(10.83)
In a mixture of gases, if all but one mole fraction is determined, the last can be
calculated from the above equation. Again, in terms of masses
p1V = m1R 1T
p2V = m2R 2T
.........................
pc V = mcR c T
348
Engineering Thermodynamics
Adding, and using Dalton’s law
pV = (m1R 1 + m2R 2 + .... + mc R c )T
where
For the gas mixture
(10.84)
p = p1 + p2 + ... + pc
pV = (m1 + m2 + ... + mc) R mT
(10.85)
where R m is the gas constant for the mixture. From Eqs. (10.84) and (10.85)
Rm =
m1 R1 + m2 R2 + ... + mc Rc
m1 + m2 + ... + mc
(10.86)
The gas constant of the mixture is thus the weighted mean, on a mass basis, of
the gas constants of the components.
The total mass of gas mixture m is
m = m1 + ... + mc
If m denotes the equivalent molecular weight of the mixture having n total number of moles.
nm = n1 m1 + n2 m 2 + ... + nc mc
\
m = x1 m1 + x2 m2 + ... + xc mc
m = S xk mK
(10.87)
A quantity called the partial volume of a component of a mixture is the volume
that the component alone would occupy at the pressure and temperature of the mixture. Designating the partial volumes by V1, V2, etc.
or
pV1 = m1R 1T, pV2 = m2R 2T, ..., pVc = mcR c T
or
p(V1 + V2 + ... + Vc ) = (m1R 1 + m2R 2 + ... + mc R c)T
(10.88)
From Eqs. (10.84), (10.85), and (10.88)
V = V1 + V2 + ... + Vc
(10.89)
The total volume is thus equal to the sum of the partial volumes.
The specific volume of the mixture, v, is given by
v=
or
m + m2 + ... + mc
1
= 1
v
V
=
or
V
V
=
m
m1 + m2 + ... + mc
m1 m2
m
+
+ ... + c
V
V
V
1
1
1
1
=
+
+ ... +
v
v1 v 2
vc
(10.90)
where v1, v2, ... denote specific volumes of the components, each component occupying the total volume.
Properties of Gases and Gas Mixtures
Therefore, the density of the mixture
r = r1 + r 2 + ... + r c
349
(10.91)
10.9 INTERNAL ENERGY, ENTHALPY AND SPECIFIC HEATS
OF GAS MIXTURES
When gases at equal pressures and temperatures are mixed adiabatically without
work, as by inter-diffusion in a constant volume container, the first law requires that
the internal energy of the gaseous system remains constant, and experiments show
that the temperature remains constant. Hence the internal energy of a mixture of
gases is equal to the sum of the internal energies of the individual components, each
taken at the temperature and volume of the mixture (i.e. sum of the ‘partial’ internal
energies). This is also true for any of the thermodynamic properties like H, C v, Cp ,
S, F, and G, and is known as Gibbs theorem. Therefore, on a mass basis
mu m = m1u1 + m2u 2 + ... + mcu c
um =
m1u1 + m2 u2 + ... + mc uc
m1 + m2 + ... + mc
(10.92)
which is the average specific internal energy of the mixture.
Similarly, the total enthalpy of a gas mixture is the sum of the ‘partial’
enthalpies
mhm = m1h1 + m2h2 + ... + mchc
and
hm =
m1 h1 + m2 h2 + ... + mc hc
m1 + m2 + ... + mc
(10.93)
From the definitions of specific heats, it follows that
cvm =
and
cpm =
m1 cv1 + m2 cv 2 + ... + mc cvc
m1 + m2 + ... + mc
m1cp + m2 c p 2 + ... + mc c pc
1
m1 + m2 + ... + mc
(10.94)
(10.95)
10.10 ENTROPY OF GAS MIXTURES
Gibbs theorem states that the total entropy of a mixture of gases is the sum of the
partial entropies. The partial entropy of one of the gases of a mixture is the
entropy that the gas would have if it occupied the whole volume alone at the same
temperature. Let us imagine a number of inert ideal gases separated from one another by suitable partitions, all the gases being at the same temperature T and pressure p. The total entropy (initial)
S i = n1s1 + n2s2 + ... + nc sc
= S nK s K
350
Engineering Thermodynamics
From property relation
Tds = dh – vdp = cp dT – vdp
\
dT
dp
-R
T
p
d s = cp
The entropy of 1 mole of the Kth gas at T and p
dT
- R ln p + s 0 K
T
where s0K is the constant of integration.
z
sK =
c pK
\
S i = R S nK
Let
sK =
1
R
z
c pK
F 1 c dT + s - ln pI
HR
K
T
R
z
0K
pK
dT s0 K
+
T
R
then
S i = R S nk (sK – ln p)
(10.96)
After the partitions are removed, the gases diffuse into one another at the same
temperature and pressure, and by Gibbs theorem, the entropy of the mixture, S f , is
the sum of the partial entropies, with each gas exerting its respective partial pressure. Thus
Since
S f = R S nK (s K – ln pK )
pK = xK p
S f = R S nK (sK – ln xK – ln p)
(10.97)
A change in entropy due to the diffusion of any number of inert ideal gases is
Sf – S i = – R S nK ln xK
(10.98)
or
Sf – S i = – R (n1 ln x 1 + n2 ln x2 + ... + nc ln xc)
Since the mole fractions are less than unity, (S f – Si ) is always positive, conforming
to the Second Law.
Again
FG
H
Sf – S i = - R n1 ln
p1
p
p
+ n2 ln 2 + ... + nc ln c
p
p
p
IJ
K
(10.99)
which indicates that each gas undergoes in the diffusion process a free expansion
from total pressure p to the respective partial pressure at constant temperature.
Similarly, on a mass basis, the entropy change due to diffusion
p
Sf – S i = – S mK R K ln K
p
FG
H
= - m1 R1 ln
p1
p
p
+ m2 R2 ln 2 + ... + mc Rc ln c
p
p
p
IJ
K
351
Properties of Gases and Gas Mixtures
10.11 GIBBS FUNCTION OF A MIXTURE OF INERT IDEAL
GASES
From the equations
dh = c p dT
ds = c p
dT
dp
-R
T
p
the enthalpy and entropy of 1 mole of an ideal gas at temperature T and pressure
p are
h = h0 + cp dT
z
dT
+ s 0 - R ln p
T
Therefore, the molar Gibbs function
s =
z
cp
g = h – Ts
= h0 +
z
c p dT - T
z
cp
dT
- T s 0 + R T ln p
T
Ú d (uv) = Ú udv + Ú vdu = uv
Now
Let
u=
1
, v = Ú cp dT
T
F
H
1
1
1
cp dT = Ú cp dT + Ú cp dT - 2
Ú
T
T
T
Then
=Ú
dT
I dT
K
z
c p dT
1
cp dT – Ú
dT
T
T2
Ú cp dT – T Ú c p T = – T Ú
z c dT dT
p
T2
Therefore
z c dT dT – T + RT ln p
T
F h - 1 z c dT dT - s + ln pI
= RT G
z
J
g = h0 – T Ú
p
s0
2
p
2
0
H RT R
0
T
R
K
Let
f=
h0
1
RT R
z z cT dT dT - sR
p
0
2
Thus
g = R T (f + ln p)
where f is a function of temperature only.
(10.100)
(10.101)
352
Engineering Thermodynamics
Let us consider a number of inert ideal gases separated from one another at the
same T and p
G i = S nK gK
= RT S n K (f K + ln p)
After the partitions are removed, the gases will diffuse, and the partial Gibbs function of a particular gas is the value of G, if that gas occupies the same volume at the
same temperature exerting a partial pressure pK. Thus
Gf = RT S n K (f K + ln pK )
= R T S n K (fK + ln p + ln xK )
Therefore
Gf – Gi = R T S nK ln xK
(10.102)
Since xK < 1, (Gf – Gi) is negative because G decreases due to diffusion. Gibbs
function of a mixture of ideal gases at T and p is thus
G = R T S nK (f K + ln p + ln xK )
(10.103)
Solved Examples
Example 10.1 A fluid at 200 kPa and 300°C has a volume of 0.8 m3. In a
frictionless process at constant volume the pressure changes to 100 kPa. Find
the final temperature and the heat transferred (a) if fluid is air, (b) if the fluid is
steam.
Solution
(a) The fluid is air.
1
1
v=
T
p
C
2
2
V
s
Fig. 10.8
p1V1
pV
= 1 2 , T 1 = 300 + 273 = 573 K
T1
T2
T2 =
p2
50
¥ T1 =
¥ 573 = 286.5 K Ans.
p1
100
kN
¥ 0.8 m3
2
p1V1
m
m=
=
= 0.4865 kg
kJ
RT1
0.287
¥ 573 K
kgK
100
353
Properties of Gases and Gas Mixtures
Q = mcv (T2 – T 1)
= 0.4865 ¥ 0.718 (286.5 – 573) = – 100.11 kJ
(b) The fluid is steam (Fig. 10.9)
Ans.
300 °C 1
t1
1
100 kPa
p1 = 100 kPa
p
T
t1 = 300 °C
t2
v2
=v
1
2 p2 = 50 kPa
50 kPa
2
s
v
Fig. 10.9
p2 = 50 kPa
From Table A.1.1,
t2 = 81.33°C, vf = 0.00103 m3/kg
vg = 3.24 m2 /kg (state 2)
At 100 kPa, 300°C,
v1 = 2.6388 m3/kg
u1 = 2810.4 kJ/kg
v1 = v2 = vf + x2 vfg
2.6388 = 0.00103 + x2 (3.24 – 0.00103)
= 0.00103 + x2 ¥ 3.23897
x2 =
\
u2 = uf + x2 u fg = 340.42 + 0.8144 ¥ 2143.4
= 2086 kJ/kg
m=
\
2.63777
= 0.8144
3.23897
0.8
V
=
= 0.383 kg
2.6388
v1
Q = m (u2 – u 1) = 0.383 (2086 – 2810.4)
= – 277.44 kJ
Ans.
T2 = 81.33 + 273 = 354.33 K Ans.
Example 10.2 A fluid having a temperature of 150°C and a specific volume of
0.96 m3/kg at its initial state expands at constant pressure, without friction, until
the volume is 1.55 m3/kg. Find, for 1 kg of fluid, the work, the heat transferred, and
the final temperature if (a) the fluid is air, (b) the fluid is steam.
Solution
(a) The fluid is air (Fig. 10.10)
p=
0.287 (150 + 273)
RT
=
= 126.46 kPa
0.96
v
354
Engineering Thermodynamics
2
T
p
p
1
2
=c
1
s
v
Fig. 10.10
W = p (v2 – v1) = 126.46
= 74.61 kJ
kN
m
2
(1.55 – 0.96) m3
Ans.
V2
1.55
= 423 ¥
= 682.97 K
V1
0.96
T2 = T1
= 409.97°C
Ans.
Q = Cp (T 2 – T 1) = 1.005 (409.97 – 150)
= 261.27 kJ
Ans.
(b) The fluid is steam (Fig. 10.11).
2
t1
T
p
1
1
2
t
s
v
Fig. 10.11
At 150°C, vg = 0.3928 m3/kg
Since v1 > vg, the state is superheated.
From superheated steam table,
p1 = 200 kPa = p2
and
h1 = 2768.8 kJ/kg
v2 = 1.55 m3/kg
\
t2 = 400°C, h2 = 3276.5 kJ/kg
Q = h2 – h1 = 3276.5 – 2768.8 = 507.7 kJ
W = p (v2 – v1) = 200 (1.55 – 0.96)
= 118 kJ Ans.
t2 = 400°C Ans.
Ans.
355
Properties of Gases and Gas Mixtures
Example 10.3 A fluid at 250°C and 300 kPa is compressed reversibly and isothermally to 1/16th of its original volume. Find the final pressure, the workdone,
and the change of internal energy per kg of fluid, (a) if the fluid in air, (b) the fluid
is steam.
Solution
(a) The fluid is air (Fig. 10.12)
2
p
T
2
1
1
s
v
Fig. 10.12
p1v1
p v
= 2 2
T1
T2
\ p2 = p 1
W1–2 = p1v1 ln
v2
v
1
= RT 1 ln 2 = 0.287 ¥ 523 ln
16
v1
v1
v1
= 300 ¥ 16 = 4800 kPa. Ans.
v2
= – 2.773 ¥ 0.287 ¥ 523 = – 416.17 kJ
Du = u2 – u1 = 0 Ans.
(b) The fluid is steam.
From the steam tables at 300 kPa, 250°C
v1 = 0.7964 m3/kg, u1 = 2728.7 kJ/kg
v2 =
1
v1 = 0.0498 m3/kg, s1 = 7.5165 kJ/kg K
16
t=c
c.p.
p
Ans.
c.p.
t=
2
x2
T
c
2
1
1
x2
s
v
Fig. 10.13
v2 = 0.0498 m3 /kg
= (vf + x2 vfg)250°C = 0.00125 + x2 ¥ (0.05013)
x2 = 0.968, s2 = 2.7927 + 0.968 ¥ 3.2802 = 5.968 kJ/kg K
356
Engineering Thermodynamics
u 2 = 1080.37 + 0.968 ¥ 1522.0 = 2553.67 kJ/kg
Du = u2 – u1 = 2553.67 – 2728.7 = 175 kJ/kg Ans.
Q1–2 = T(s2 – s1) = (250 + 273) (5.968 – 7.5165)
= – 523 ¥ 1.5485 = – 809.87 kJ/kg
Ans.
Example 10.4 A reversible adiabatic process begins at p1 = 10 bar, t1 = 300°C
and ends with p2 = 1 bar. Find the specific volume and the work done per kg of
fluid if (a) the fluid is air, (b) the fluid is steam.
Solution
(a) The fluid is air.
1
1
pv2 = c
p
T
2
2
s
v
Fig. 10.14
p1 v1g = p2 v2g
2
W1–2 = Ú p dv =
1
=
RT1 Ê T2
ˆ
- 1˜
1 - g ÁË T1
¯
g -1
T2
Êp ˆ g
= Á 2˜
T1
Ë p1 ¯
\
W1–2 =
p1v1 - p2 v2
R
=
(T – T 2)
g -1
g -1 1
0.287 ¥ 573
(0.519 – 1)
1 - 1.4
= 197.75 kJ/kg
v2 =
1.4 - 1
1
Ê 1 ˆ 1.4
= Á ˜
=
= 0.519
Ë 10 ¯
1.93
Ans.
0.287 ¥ 0.519 ¥ 573
RT2
=
100
p2
= 0.854 m 3/kg Ans.
(b) The fluid is steam (Fig. 10.15)
u 1 = 2793.2 kJ/kg, v1 = 0.2579 m3/kg, s1 = 7.1228
s1 = s2 = sf + x2 sfg
kJ
kg K
357
Properties of Gases and Gas Mixtures
1
1
10 bar
p
t = 300°C
10 bar
s=
T
c
1 bar
1 bar
2
2
s
v
Fig. 10.15
7.1228 = 1.3025 + x2 ¥ 6.0568
5.8203
= 0.96
6.0568
u 2 = 417.33 + 0.96 ¥ 2088.7 = 2422.5 kJ/g
v2 = 0.001043 + 0.96 ¥ 1.693 = 1.626 m3/kg
W1–2 = u1 – u 2 = 2793.2 – 2422.5
= 370.7 kJ/kg Ans.
x2 =
Ans.
Example 10.5 A reversible polytropic process begins with a fluid at p1 = 10
bar, T1 = 200°C and ends at p2 = 1 bar. The exponent n has the value 1.15. Find the
final specific volume, the final temperature and the heat transferred per kg of fluid
if (a) the fluid is air, (b) the fluid is steam.
Solution
Fig. 10.16
(a) The fluid is air (Fig. 10.16).
1
1
0.287 ¥ 473
RT1 Ê p1 ˆ n
Ê p ˆn
(10)1/1.15
v 2 = v1 Á 1 ˜ =
=
1000
Ë p2 ¯
p1 ÁË p2 ˜¯
v2 = 0.13575 ¥ 7.406 = 1.0054 m3/kg
p1v1
p v
= 2 2
T1
T2
Ans.
358
\
Engineering Thermodynamics
T 2 = 473 ¥
1
1.0054
= 350.306 K = 77.3°C Ans.
¥
10 0.13575
Q = Du + W = cv DT +
p2 v2 - p1v1
1- n
R ˆ
Ê
= Á cv +
(T2 – T1)
1 - n ˜¯
Ë
0.287 ˆ
Ê
= ÁË 0.716 +
˜ (350.306 – 473)
-0.15 ¯
= (– 1.195) (– 123.306) = 147.35 kJ
(b) The fluid is steam (Fig. 10.17)
Ans.
Fig. 10.17
At 10 bar, 200°C,
v1 = 0.20596, u1 = 2621.9
1
1
Ê p ˆn
v2 = v1 Á 1 ˜ = 0.20596 ¥ (10)1.15
Ë p2 ¯
p2 = 0.20596 ¥ 7.406 = 1.525 m3/kg
Ans.
3
vg at 0.1 MPa,vg = 1.694 m /kg
Since v2 < vg, the state 2 is in the mixture region.
v2 = 1.525 = 0.001043 + x2 ¥ (1.694 – 0.001043)
x2 = 0.933
t2 = (t sat)1 bar = 99.62°C Ans.
u 2 = 417.33 + 0.933 ¥ 2088.7 = 2366.1 kJ/kg
2
W = Ú pdv =
1
=
p1v1 - p2 v2
1000 ¥ 0.13575 - 100 ¥ 1.525
=
n -1
0.15
135.75 - 152.5
= – 111.67 kJ/kg
0.15
359
Properties of Gases and Gas Mixtures
Q = Du + W = u 2 – u1 + W
= (2366.1 – 2621.9) – 111.67
= – 255.8 – 111.67 = – 367.47 kJ/kg
Ans.
Example 10.6 An ideal monatomic gas occupies a volume of 10–3 m3 at temperature 3 K and pressure 103 Pa. The internal energy of the gas is taken to be zero
at this point. It undergoes the following cycle.
The temperature is raised to 300 K at constant volume, the gas is expanded
adiabatically till the temperature is 3 K, followed by isothermal compression to the
original volume. Plot on a p-V diagram. Calculate (a) the work done and the heat
transferred in each process and the internal energy at the end of each process,
(b) the thermal efficiency of the cycle.
Solution
pV = nRT
At initial conditions
(103 ¥ 10–3) J = nR ¥ 3
1
= number of moles of the gas.
3R
\
h=
(a) Process ab
Here
dV = 0, Wab = 0
Qab = ncv (Tb – Ta)
For a monatomic gas, cv =
3
R
2
1
3
¥ R ¥ 297 = 148.5 J
3R 2
Ub – Ua = 148.5 J
Qab =
But
Fig. 10.18
Ua = 0 (given)
Ub = 148.5 J
Ans.
Process bc
Qbc = 0, Wbc = – DU = – (UC – Ub)
Since T C = 3 K and U is a function of the temperature only and U = 0 at T = 3 K
(given)
\
UC = 0 and Wbc = Ub = 148.5 J Ans.
Process ca
Wca = nRT ln
Va
V
= – nRT ln c
Vc
Vb
For the adiabatic process bc,
TV g – 1 = C
and
g =
5
3
360
Engineering Thermodynamics
Ê Vc ˆ
ÁË V ˜¯
b
2/3
Tb
300
=
= 100 = 102
3
Tc
=
Vc
= 10 3
Vb
\
1
◊ R ◊ 3 ◊ ln 103 = – 6.9 J
3R
Qca = – 6.9 J and Ua = UC = 0
Wca = –
(b) h =
148.5 - 6.9
= 0.954
148.5
Ans.
Ans.
Example 10.7 Two vessels, A and B, both containing nitrogen, are connected
by a valve which is opened to allow the contents to mix and achieve an equilibrium
temperature of 27°C. Before mixing the following information is known about the
gases in the two vessels.
Vessel A
Vessel B
p = 1.5 MPa
t = 50°C
Contents = 0.5 kg mol
p = 0.6 MPa
t = 20°C
Contents = 2.5 kg
Calculate the final equilibrium pressure, and the amount of heat transferred to
the surroundings. If the vessel had been perfectly insulated, calculate the final
temperature and pressure which would have been reached. Take g = 1.4
Solution For the gas in vessel A (Fig. 10.19)
where VA is the volume of vessel A
1.5 ¥ 10 3 ¥ VA = 0.5 ¥ 8.3143 ¥ 323
3
VA = 0.895 m
The mass of gas in vessel A
mA = nA m A
A
1.5 MPa
50ºC
0.5 kg mole
323 K
= 0.5 kg mol ¥ 28 kg/kg mol = 14 kg
Characteristic gas constant R of nitrogen
8.3143
R=
= 0.297 kJ/kg K
28
For the vessel B
pB VB = mB RT B
B
0.6 MPa
20ºC
2.5 kg
293 K
Fig. 10.19
0.6 ¥ 10 3 ¥ VB = 2.5 ¥ 0.297 ¥ 293
\
VB = 0.363 m3
Total volume of A and B
V = VA + VB = 0.895 + 0.363 = 1.258 m3
Properties of Gases and Gas Mixtures
361
Total mass of gas
m = mA + mB = 14 + 2.5 = 16.5 kg
Final temperature after mixing
T = 27 + 273 = 300 K
For the final condition after mixing
pV = mRT
where p is the final equilibrium pressure
\
p ¥ 1.258 = 16.5 ¥ 0.297 ¥ 300
\
p=
16.5 ¥ 0.297 ¥ 300
= 1168.6 kPa = 1.168 MPa
1.258
0.297
R
=
= 0.743 kJ/kg K
0.4
g -1
Since there is no work transfer, the amount of heat transfer
Q = change of internal energy = U2 – U1
Measuring the internal energy above the datum of absolute zero (at T = 0 K,
u = 0 kJ/kg)
Initial internal energy U1 (before mixing)
= mA cvT A + mB c vTB
cv =
= (14 ¥ 323 + 2.5 ¥ 293) ¥ 0.743 = 3904.1 kJ
Final internal energy U2 (after mixing)
= mcvT
= 16.5 ¥ 0.743 ¥ 300 = 3677.9 kJ
\
Q = 3677.9 – 3904.1 = – 226.2 kJ
If the vessels were insulated
U1 = U2
mA c vTA + mB cvT B = mc vT
where T would have been the final temperature.
m T + m B TB
\
T= A A
m
14 ¥ 323 + 2.5 ¥ 293
=
= 318.5 K
16.5
or
t = 45.5°C
The final pressure
16.5 ¥ 0.297 ¥ 318.5
mRT
p=
=
= 1240.7 kPa = 1.24 MPa
1.258
V
Example 10.8 A certain gas has cp = 1.968 and c v = 1.507 kJ/kg K. Find its
molecular weight and the gas constant.
A constant volume chamber of 0.3 m3 capacity contains 2 kg of this gas at 5°C.
Heat is transferred to the gas until the temperature is 100°C. Find the work done,
the heat transferred, and the changes in internal energy, enthalpy and entropy.
362
Engineering Thermodynamics
Gas constant,
R = cp – cv = 1.968 – 1.507 = 0.461 kJ/kg K
Molecular weight,
8.3143
R
m=
=
= 18.04 kg/kg mol
0 .461
R
At constant volume
Q1 – 2 = mcv (t2 – t1)
Solution
= 2 ¥ 1.507 (100 – 5) = 286.33 kJ
W1 – 2 =
z
2
1
pdv = 0
Change in internal energy
U2 – U1 = Q1 – 2 = 286.33 kJ
Change in enthalpy
H2 – H1 = mcp (t2 – t1)
= 2 ¥ 1.968 (100 – 5) = 373.92 kJ
Change in entropy
T2
373
= 2 ¥ 1.507 ln
= 0.921 kJ/K
268
T1
Example 10.9 (a) The specific heats of a gas are given by cp = a + kT and c v =
b + kT, where a, b, and k are constants and T is in K. Show that for an isentropic
expansion of this gas
T b va–b e kT = constant
(b) 1.5 kg of this gas occupying a volume of 0.06 m3 at 5.6 MPa expands isentropically until the temperature is 240°C. If a = 0.946, b = 0.662, and k = 10–4,
calculate the work done in the expansion.
S2 – S1 = mcv ln
Solution
(a) cp – cv = a + kT – b – kT
=a–b=R
dT
dv
+R
T
v
dT
dv
dT
dv
= (b + kT )
+ (a – b)
=b
+ k dT + (a – b)
T
v
T
v
For an isentropic process
Now
ds = cv
b ln T + kT + (a – b) ln v = constant
\
T b ◊ va – b ekT = constant
(Q.E.D.)
(b) R = a – b = 0.946 – 0.662 = 0.284 kJ/kg K
T2 = 240 + 273 = 513 K
T1 =
5.6 ¥ 10 3 ¥ 0.06
p1V1
=
= 788.73 K = 789 K
1.5 ¥ 0.284
mR
363
Properties of Gases and Gas Mixtures
TdS = dU + d- W = 0
W1 – 2 = -
\
z
T2
mcv dT
T1
= 1.5
z
789
513
(0.662 + 0.0001T ) dT
= 1.5 [0.662 (789 – 513) + 10 –4 ¥ 0.5 {(789)2 – (513)2}]
= 1.5 (182.71 + 19.97) = 304 kJ
Example 10.10 Show that for an ideal gas, the slope of the constant volume
line on the T–s diagram is more than that of the constant pressure line.
Solution
We have, for 1 kg of ideal gas
Tds = du + pdv
v=c
= cv dT + pdv
v
Also
p=c
FG ∂T IJ = T
H ∂s K c
v
T
\
Tds = dh – vdp
= cp dT – vdp
\
p
Since
\
A
FG ∂T IJ = T
H ∂s K c
p
cp > cv ,
T
T
>
cv c p
FG ∂T IJ > FG ∂T IJ
H ∂s K H ∂s K
v
s
Fig. 10.20
p
This is shown in Fig. 10.20. The slope of the constant volume line passing through
point A is steeper than that of the constant pressure line passing through the same
point.
(Q.E.D.)
Example 10.11 0.5 kg of air is compressed reversibly and adiabatically from
80 kPa, 60°C to 0.4 MPa, and is then expanded at constant pressure to the original
volume. Sketch these processes on the p–v and T–s planes. Compute the heat transfer and work transfer for the whole path.
Solution The processes have been shown on the p–v and T–s planes in Fig. 10.21.
At state 1
p1V1 = mRT1
\
V1 = volume of air at state 1
=
mRT1 1 ¥ 0.287 ¥ 333
=
= 0.597 m3
p1
2 ¥ 80
364
Engineering Thermodynamics
v=c
p=c
3
3
p
T
2
g
pV = C
2
1
1
s
v
(a)
(b)
Fig. 10.21
Since the process 1–2 is reversible and adiabatic
( g -1)/ g
FG p IJ
HpK
T
F 400 I
=
H 80 K
T
T2
=
T1
\
2
1
(1. 4 -1)/1. 4
2
= (5)2/7
1
\
T2 = 333 ¥ (5)2/7 = 527 K
For process 1–2, work done
p V - p2 V2
mR (T1 - T2 )
W1–2 = 1 1
=
g -1
g -1
1/ 2 ¥ 0.287 (333 - 527)
= – 69.6 kJ
0.4
p1 v g1 = p2 vg2
=
Again
g
\
FG v IJ = p = 80 = 1
H v K p 400 5
v
F 1I = 1 = V
=
H 5K 3.162 V
v
2
1
1
2
1/1. 4
\
\
2
2
1
1
V2 =
0.597
= 0.189 m3
3.162
For process 2–3, work done
W2–3 = p2 (V1 – V2) = 400 (0.597 – 0.189) = 163.2 kJ
\ Total work transfer
W = W1–2 + W2–3
= – 69.6 + 163.2 = 93.6 kJ
For states 2 and 3
p2 V2
pV
= 3 3
T3
T2
365
Properties of Gases and Gas Mixtures
\
T3 = T2 ◊
V3
= 527 ¥ 3.162 = 1667 K
V2
Total heat transfer
Q = Q1 – 2 + Q2 – 3 = Q2 – 3 = mc p (T 3 – T 2)
= 1/2 ¥ 1.005 (1667 – 527) = 527.85 kJ
Example 10.12 A mass of air is initially at 260°C and 700 kPa, and occupies
0.028 m3. The air is expanded at constant pressure to 0.084 m3. A polytropic process with n = 1.50 is then carried out, followed by a constant temperature process
which completes a cycle. All the processes are reversible. (a) Sketch the cycle in
the p–v and T–s planes. (b) Find the heat received and heat rejected in the cycle.
(c) Find the efficiency of the cycle.
(a) The cycle is sketched on the p–v and T–s planes in Fig. 10.22.
2
pV 1.5 = C
p
1
2
T
Solution
p
pV = C
=
1
3
pV 1.5 = C
C
3
s
V
(a)
(b)
Fig. 10.22
Given
p1 = 700 kPa, T1 = 260 + 273 = 533 K = T 3
V1 = 0.028 m3
V2 = 0.084 m3
From the ideal gas equation of state
p1V1 = mRT1
m=
700 ¥ 0.028
= 0.128 kg
0.287 ¥ 533
Now
0.084
T2
pV
= 2 1 =
=3
0.028
T1
p1V1
\
T2 = 3 ¥ 533 = 1599 K
Again
p2
T
= 2
p3
T3
FG IJ
H K
n /( n -1)
=
F 1.5 / 0.5 I
H 533 K
1. 5 / 0 . 5
= (3)3 = 27
366
Engineering Thermodynamics
Heat transfer in process 1–2
Q1–2 = mcp (T2 – T 1)
= 0.128 ¥ 1.005 (1599 – 533) = 137.13 kJ
Heat transfer in process 2–3
Q2 – 3 = DU + Ú pdv
= m c v (T 3 – T 2) +
= mcv
mR (T2 - T3 )
n -1
1.5 - 1.4
n-g
(T 3 – T2) = 0.128 ¥ 0.718 ¥
(533 – 1599)
1.5 - 1
n -1
= 0.128 ¥ 0.718 ¥
0.1
(– 1066) = – 19.59 kJ
0.5
For process 3–1
d- Q = dU + d- W = d- W
1
\
Q 3 – 1 = W3 – 1 =
= mRT1 ln
z
3
pdV = mRT 1 ln
V1
V3
F I
H K
p3
1
= 0.128 ¥ 0.287 ¥ 533 ln
p1
27
= – 0.128 ¥ 0.287 ¥ 533 ¥ 3.2959 = – 64.53 kJ
(b) Heat received in the cycle
Q1 = 137.13 kJ
Heat rejected in the cycle
Q2 = 19.59 + 64.53 = 84.12 kJ
(c) The efficiency of the cycle
hcycle = 1 –
Q2
84.12
=1–
= 1 – 0.61 = 0.39, or 39%
137.13
Q1
Example 10.13 A mass of 0.25 kg of an ideal gas has a pressure of 300 kPa, a
temperature of 80°C, and a volume of 0.07 m3. The gas undergoes an irreversible
adiabatic process to a final pressure of 300 kPa and final volume of 0.10 m3, during which the work done on the gas is 25 kJ. Evaluate the cp and cv of the gas and
the increase in entropy of the gas.
Solution
From
p1V1 = mRT1
R=
300 ¥ 0.07
= 0.238 kJ/kg K
0.25 ¥ (273 + 80 )
T2 =
300 ¥ 0.1
p2 V2
=
= 505 K
0.25 ¥ 0.238
mR
Final temperature
367
Properties of Gases and Gas Mixtures
Now
Q = (U2 – U1) + W = mc v (T 2 – T1) + W
0 = 0.25 c v (505 – 353) – 25
25
= 0.658 kJ/kg K
0.25 ¥ 152
\
cv =
Now
cp – cv = R
cp = 0.658 + 0.238 = 0.896 kJ/kg K
Entropy change
S2 – S1 = mcv ln
= mcp ln
p2
v
+ mc p ln 2
v1
p1
V2
0.10
= 0.25 ¥ 0.896 ln
0.07
V1
= 0.224 ¥ 0.3569 = 0.08 kJ/kg K
Example 10.14 Atmospheric air is sucked in an internal combustion engine
cylinder. The pressure, temperature and volume of air after suction are 1 bar, 75°C
and 800 cm3 respectively. The air is then reversibly compressed to 15 bar and 1/8th
of its volume. Heat of combustion of fuel is then added to air at constant volume
until its pressure is 50 bar. (a) Determine the index of compression process. (b)
Calculate the change of entropy of air during each process. (c) Determine if there
is any heat exchange between air and cylinder walls during compression. State the
direction of heat flow.
Solution The processes are shown on p–V and T-s diagrams in Fig. 10.23
(a) Index of compression process
ln p2 / p1
ln 15
n=
=
= 1.3
ln V1 / V2
ln 8
3
3
p
v
=
C
p2 = 15 bar
pV n = C
T
2
2
p1 = 1 bar
1
0.0001 m3
pV n = C
1
0.0008 m3
V
S
(a)
(b)
Fig. 10.23
368
Engineering Thermodynamics
(b) Mass of air in the cylinder
pV
100 ¥ 800 ¥ 10 -6
m= 1 1 =
= 0.0008 kg
0.287 ¥ 348
RT1
p2 V2
pV
= 1 1
T2
T1
pV
1
\
T2 = T1 2 2 = 348 ¥ 15 ¥ = 652.5 K
8
p1V1
p3
50
T3 = T2 ◊
= 652.5 ¥
= 2175 K
15
p2
LM
N
T
V
S2 – S1 = m c2 ln 2 + R ln 2
T1
V1
LM
N
OP
Q
OP
Q
652.5
1
+ 0.287 ln
348
8
= 0.0008 [0.4513 – 0.5968] = – 0001164 kJ/K
= 0.0008 0.718 ln
2175
T3
= 0.0008 ¥ 0.718 ln
= 0.000692 kJ/K
652.5
T2
(c) Heat transfer during polytropic compression process
g -n
Q = mcv
(T2 – T1 )
1- n
1.4 - 1.3
= 0.0008 ¥ 0.718
(652.5 – 348) = – 0.058 kJ
1 - 1.3
The direction of heat flow is from air to cylinder walls, which is thus negative.
S3 – S2 = m c v ln
Example 10.15 A mixture of ideal gases consists of 3 kg of nitrogen and 5 kg of
carbondioxide at a pressure of 300 kPa and a temperature of 20°C. Find (a) the
mole fraction of each constituent, (b) the equivalent molecular weight of the
mixture, (c) the equivalent gas constant of the mixture, (d) the partial pressures
and the partial volumes, (e) the volume and density of the mixture, and (f ) the cp
and cv of the mixture.
If the mixture is heated at constant volume to 40°C, find the changes in internal
energy, enthalpy and entropy of the mixture. Find the changes in internal energy,
enthalpy and entropy of the mixture if the heating is done at constant pressure.
Take g for CO2 and N2 to be 1.286 and 1.4 respectively.
Solution
(a) Since mole fraction
xi =
xN2 =
ni
S ni
3
28
3
5
+
28 44
= 0.485
Properties of Gases and Gas Mixtures
369
5
44
xCO2 =
= 0.515
3
5
+
28 44
(b) Equivalent molecular weight of the mixture
M = x1 m1 + x 2 m2
= 0.485 ¥ 28 + 0.515 ¥ 44 = 36.25 kg/kg mol
(c) Total mass,
m = mN2 + mCO2 = 3 + 5 = 8 kg
Equivalent gas constant of the mixture
R=
mN2 RN2 + mCO2 RCO2
m
8.3143
8.3143
+5¥
28
44 = 0.89 + 0.94 = 0.229 kJ/kg K
=
8
8
(d) pN2 = xN2 ◊ p = 0.485 ¥ 300 = 145.5 kPa
pCO2 = xCO2 ◊ p = 0.515 ¥ 300 = 154.5 kPa
3¥
VN2 =
VCO2 =
3¥
mN2 RN2 T
=
p
mCO2 RCO2 T
8.3143
¥ 293
28
= 0.87 m3
300
5¥
8.3143
¥ 293
44
= 0.923 m3
300
=
p
(e) Total volume of the mixture
\
V=
mN2 RN2 T
mCO2 RCO2 T
mRT
=
=
pN 2
pCO2
p
V=
8 ¥ 0.229 ¥ 293
= 1.79 m3
300
Density of the mixture
r = rN2 + rCO2 =
(f ) cpN2 – cvN2 = RN2
\
cvN2 =
RN2
g -1
=
8
m
=
= 4.46 kg/m3
1.79
V
8.3143
28 ¥ (1.4 - 1)
– 0.742 kJ/kg K
\
For CO2,
cpN2 = 1.4 ¥ 0.742 = 1.039 kJ/kg K
g = 1.286
370
Engineering Thermodynamics
\
cvCO2 =
RCO2
=
g -1
8.3143
= 0.661 kJ/kg K
44 ¥ 0.286
cpCO2 = 1.286 ¥ 0.661 = 0.85 kJ/kg K
For the mixture
cp =
mN 2 c pN2 + mCO 2 c pCO 2
mN2 + mCO 2
= 3/8 ¥ 1.039 + 5/8 ¥ 0.85 = 0.92 kJ/kg K
cv =
mN2 cvN2 + mCO2 cvCO2
m
= 3/8 ¥ 0.742 + 5/8 ¥ 0.661 = 0.69 kJ/kg K
If the mixture is heated at constant volume
U2 – U1 = mcv (T2 – T1) = 8 ¥ 0.69 ¥ (40 – 20) = 110.4 kJ
H2 – H1 = mcp (T2 – T1) = 8 ¥ 0.92 ¥ 20 = 147.2 kJ
T2
V
+ mR ln 2
T1
V1
T2
313
= mcv ln
= 8 ¥ 0.69 ¥ ln
= 0.368 kJ/kg K
293
T1
If the mixture is heated at constant pressure, DU and DH will remain the same.
The change in entropy will be
T
p
S2 – S1 = mcp ln 2 – mR ln 2
p1
T1
T
313
= mcp ln 2 = 8 ¥ 0.92 ln
= 0.49 kJ/kg K
293
T1
S2 – S1 = mcv ln
Example 10.16 Find the increase in entropy when 2 kg of oxygen at 60°C are
mixed with 6 kg of nitrogen at the same temperature. The initial pressure of each
constituent is 103 kPa and is the same as that of the mixture.
2
pO2
32
xO2 =
=
= 0.225
p
2
6
+
32 28
xN2 =
pN2
= 0.775
p
Entropy increase due to diffusion
pO2
pN2
DS = – mO2 RO2 ln
– mN2 RN2 ln
p
p
=–2
F 8.3143 I ln 0.225 – 6 F 8.3143 I ln 0.775 = 1.2314 kJ/kg K
H 32 K
H 28 K
371
Properties of Gases and Gas Mixtures
Example 10.17 The gas neon has a molecular weight of 20.183 and its critical
temperature, pressure and volume are 44.5 K, 2.73 MPa and 0.0416 m3/kg mol.
Reading from a compressibility chart for
Z = 0.7
Tr = 1.3
a reduced pressure of 2 and a reduced
temperature of 1.3, the compressibility
factor Z is 0.7. What are the correspondZ
ing specific volume, pressure, temperature, and reduced volume?
Solution At pr = 2 and Tr = 1.3 from
chart (Fig. 10.24)
Z = 0.7
p = 2 ¥ 2.73 = 5.46 MPa
0.7
0
2
T
= 1.3
Tc
pr
Fig. 10.24
T = 1.3 ¥ 44.5 = 57.85 K
pv = ZRT
\
v=
0.7 ¥ 8.3143 ¥ 57.85
= 3.05 ¥ 10–3 m3/kg
20.183 ¥ 5.46 ¥ 10 3
\
vr =
v
3.05 ¥ 10 -3 ¥ 20.183
=
= 1.48
vc
4.16 ¥ 10 -2
Example 10.18
For the Berthelot equation of state
p=
RT
a
- 2
v - b Tv
show that (a) lim (RT – pv) = 0
p Æ0
T Æ•
R
v
= ,
T Æ• T
p
(b) lim
(c) Boyle temperature, TB =
a
,
bR
1 2aR
, vc = 3b, Tc =
12b 3b
(e) Law of corresponding states
(d) Critical properties p c =
Ê
3 ˆ
Á pr +
˜ (3vr – 1) = 8Tr
Tr × vr2 ¯
Ë
8a
,
27bR
372
Solution
Engineering Thermodynamics
(a) p =
RT
a
- 2
v - b Tv
F
H
a
Tv 2
I (v – b)
K
\
RT = p +
or
RT
a
ab
=v+
–b–
p
pvT
pv 2 T
\
RT – pv =
\
a
ab
– bp – 2
vT
v T
lim (RT – pv) = 0
Proved (a)
pÆ0
T Æ•
RT
a
ab
+b+
p
pvT
pv 2 T
(b) Now
v=
\
v
R
a
b
ab
+ + 2 2
= 2
p
T
pvT
T pv T
\
lim
v
T Æ• T
=
R
p
Proved (b)
a
ab
+ bp + 2
vT
v T
The last three terms of the equation are very small, except at very high pressures
and small volume. Hence substituting v = RT/p
(c)
pv = RT –
pv = RT –
ap
abp 2
+
bp
+
RT 2
R2 T 3
LM ∂ ( pv) OP = - a + b + 2 abp = 0
R T
N ∂p Q RT
\
2
2
3
T
when
p = 0, T = TB, the Boyle temperature
\
a
=b
RTB2
or
TB =
(d)
p=
a
bR
RT
a
- 2
v - b Tv
FG ∂p IJ
H ∂v K
= T = Tc
2a
RTc
+
=0
( v c - b ) 2 Tc ◊ v c3
Proved (c)
Properties of Gases and Gas Mixtures
373
FG ∂ p IJ = 2 RT - 6 a = 0
(v - b)
T ◊v
H ∂v K
F p + a I (v – b) = RT
GH T v JK
2
c
2
c
3
T = Tc
c
c
2
c c
c
c
4
c
By solving the three equations, as was done in the case of van der Waals
equation of state in Article 10.7
1
2 aR
, vc = 3b, and Tc =
12 b 3b
(e) Solving the above three equations
pc =
8a
27bR
a=
8v c3 pc2
= 3pc ◊ v 2c ◊ Tc
R
b=
vc
8 pv v c
,R=
(so that Zc = 3/8)
3 Tc
3
Proved (d)
Substituting in the equation
F p + a I (v – b) = RT
H Tv K
F p + 3p v T I Fv - v I = 8 p v ◊ T
GH Tv JK H 3 K 3T
FG p + 3 IJ = (3v – 1) = 8T
H Tv K
2
2
c c c
2
c
c c
c
r
2
r r
r
r
This is the law of corresponding states.
Proved (e)
Review Questions
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8
10.9
What is a mole?
What is Avogadro’s law?
What is an equation of state?
What is the fundamental property of gases with respect to the product pv?
What is universal gas constant?
Define an ideal gas.
What is the characteristic gas constant?
What is Boltzmann constant?
Why do the specific heats of an ideal gas depend only on the atomic structure
of the gas?
10.10 Show that for an ideal gas the internal energy depends only on its temperature.
10.11 Show that the enthalpy of an ideal gas is a function of temperature only.
10.12 Why is there no temperature change when an ideal gas is throttled?
10.13 Show that for an ideal gas, cp – cv = R.
374
Engineering Thermodynamics
10.14 Derive the equations used for computing the entropy change of an ideal gas.
10.15 Show that for a reversible adiabatic process executed by an ideal gas, the
following relations hold good: (i) pvg = constant, (ii) Tvg – 1 = constant, and
(iii) Tp(1 – g )/g = constant.
10.16 Express the changes in internal energy and enthalpy of an ideal gas in a reversible adiabatic process in terms of the pressure ratio.
10.17 Derive the expression of work transfer for an ideal gas in a reversible isothermal process.
10.18 What is a polytropic process? What are the relations among p, v and T of an
ideal gas in a polytropic process?
10.19 Show that the entropy change between states 1 and 2 in a polytropic process,
pvn = constant, is given by the following relations:
n-g
T
(i) s2 – s1 =
R ln 2
T1
(g - 1)( n - 1)
(ii) s2 – s1 =
p
n -g
R ln 2
p1
n (g - 1)
(iii) s2 – s1 = –
v
n -g
R ln 2
v1
g -1
10.20 What are the expressions of work transfer for an ideal gas in a polytropic
process, if the gas is: (i) a closed system, and (ii) a steady system?
10.21 Derive the expression of heat transfer for an ideal gas in a polytropic process.
What is the polytropic specific heat? What would be the direction of heat transfer if (a) n > g , and (b) n < g ?
10.22 Why is the external work supplied to a compressor equal to -
z
p2
v dp?
p1
10.23 Why does isothermal compression need minimum work and adiabatic compression maximum work?
10.24 What is the isothermal efficiency of a compressor?
10.25 Write down the van der Waals equation of state. How does it differ from the
ideal gas equation of state. What is force of cohesion? What is co-volume?
10.26 What are the two-constant equations of state?
10.27 Give the virial expansions for pv in terms of p and v.
10.28 What are virial coefficients? When do they become zero?
10.29 What is the compressibility factor?
10.30 What are reduced properties?
10.31 What is the generalized compressibility chart?
10.32 What is the law of corresponding states?
10.33 Express the van der Waals constants in terms of critical properties.
10.34 Draw the diagram representing the law of corresponding states in reduced coordinates indicating the isotherms and the liquid and vapour phases.
10.35 Define Boyle temperature? How is it computed?
10.36 State Dalton’s law of partial pressures.
10.37 How is the partial pressure in a gas mixture related to the mole fraction?
10.38 How are the characteristic gas constant and the molecular weight of a gas
mixture computed?
Properties of Gases and Gas Mixtures
375
10.39 What is Gibb’s theorem?
10.40 Show that in a diffusion process a gas undergoes a free expansion from the
total pressure to the relevant partial pressure.
10.41 Show that in a diffusion process at constant temperature the entropy increases
and the Gibbs function decreases.
Problems
10.1 What is the mass of air contained in a room 6 m ¥ 9 m ¥ 4 m if the pressure is
101.325 kPa and the temperature is 25°C?
Ans. 256 kg
10.2 The usual cooking gas (mostly methane) cylinder is about 25 cm in diameter
and 80 cm in height. It is changed to 12 MPa at room temperature (27°C).
(a) Assuming the ideal gas law, find the mass of gas filled in the cylinder.
(b) Explain how the actual cylinder contains nearly 15 kg of gas. (c) If the
cylinder is to be protected against excessive pressure by means of a fusible
plug, at what temperature should the plug melt to limit the maximum pressure
to 15 MPa?
10.3 A certain gas has cp = 0.913 and cv = 0.653 kJ/kg K. Find the molecular weight
and the gas constant R of the gas.
10.4 From an experimental determination the specific heat ratio for acetylene (C2H2 )
is found to 1.26. Find the two specific heats.
10.5 Find the molal specific heats of monatomic, diatomic, and polyatomic gases, if
their specific heat ratios are respectively 5/3, 7/5 and 4/3.
10.6 A supply of natural gas is required on a site 800 m above storage level. The gas
at – 150°C, 1.1 bar from storage is pumped steadily to a point on the site where
its pressure is 1.2 bar, its temperature 15°C, and its flow rate 1000 m3/hr. If the
work transfer to the gas at the pump is 15 kW, find the heat transfer to the gas
between the two points.Neglect the change in K.E. and assume that the gas has
the properties of methane (CH4) which may be treated as an ideal gas having
g = 1.33 (g = 9.75 m/s2).
Ans. 63.9 kW
10.7 A constant volume chamber of 0.3 m3 capacity contains 1 kg of air at 5°C. Heat
is transferred to the air until the temperature is 100°C. Find the work done, the
heat transferred, and the changes in internal energy, enthalpy and entropy.
10.8 One kg of air in a closed system, initially at 5°C and occupying 0.3 m3 volume,
undergoes a constant pressure heating process to 100°C. There is no work other
than pdv work. Find (a) the work done during the process, (b) the heat transferred, and (c) the entropy change of the gas.
10.9 0.1 m3 of hydrogen initially at 1.2 MPa, 200°C undergoes a reversible isothermal expansion to 0.1 MPa. Find (a) the work done during the process, (b) the
heat transferred, and (c) the entropy change of the gas.
10.10 Air in a closed stationary system expands in a reversible adiabatic process from
0.5 MPa, 15°C to 0.2 MPa. Find the final temperature, and per kg of air, the
change in enthalpy, the heat transferred, and the work done.
10.11 If the above process occurs in an open steady flow system, find the final
temperature, and per kg of air, the change in internal energy, the heat transferred, and the shaft work. Neglect velocity and elevation changes.
376
Engineering Thermodynamics
10.12 The indicator diagram for a certain water-cooled cylinder and piston air compressor shows that during compression pv1.3 = constant. The compression starts
at 100 kPa, 25°C and ends at 600 kPa. If the process is reversible, how much
heat is transferred per kg of air?
10.13 An ideal gas of molecular weight 30 and g = 1.3 occupies a volume of 1.5 m3 at
100 kPa and 77°C. The gas is compressed according to the law pv 1.25 = constant to a pressure of 3 MPa. Calculate the volume and temperature at the end
of compression and heating, work done, heat transferred, and the total change
of entropy.
10.14 Calculate the change of entropy when 1 kg of air changes from a temperature
of 330 K and a volume of 0.15 m3 to a temperature of 550 K and a volume of
0.6 m3 .
If the air expands according to the law, pvn = constant, between the same end
states, calculate the heat given to, or extracted from, the air during the expansion, and show that it is approximately equal to the change of entropy multiplied by the mean absolute temperature.
10.15 0.5 kg of air, initially at 25°C, is heated reversibly at constant pressure until the
volume is doubled, and is then heated reversibly at constant volume until the
pressure is doubled. For the total path, find the work transfer, the heat transfer,
and the change of entropy.
10.16 An ideal gas cycle of three processes uses Argon (Mol. wt. 40) as a working
substance. Process 1–2 is a reversible adiabatic expansion from 0.014 m3 ,
700 kPa, 280°C to 0.056 m3. Process 2–3 is a reversible isothermal process.
Process 3–1 is a constant pressure process in which heat transfer is zero. Sketch
the cycle in the p-v and T-s planes, and find (a) the work transfer in process
1–2, (b) the work transfer in process 2–3, and (c) the net work of the cycle.
Take g = 1.67.
10.17 A gas occupies 0.024m3 at 700 kPa and 95°C. It is expanded in the non-flow
process according to the law pv 1.2 = constant to a pressure of 70 kPa after
which it is heated at constant pressure back to its original temperature. Sketch
the process on the p-v and T-s diagrams, and calculate for the whole process
the work done, the heat transferred, and the change of entropy. Take cp = 1.047
and cv = 0.775 kJ/kg K for the gas.
10.18 0.5 kg of air at 600 kPa receives an addition of heat at constant volume so that
its temperature rises from 110°C to 650°C. It then expands in a cylinder
polytropically to its original temperature and the index of expansion is 1.32.
Finally, it is compressed isothermally to its original volume. Calculate (a) the
change of entropy during each of the three stages, (b) the pressures at the end of
constant volume heat addition and at the end of expansion. Sketch the processes on the p-v and T-s diagrams.
10.19 0.5 kg of helium and 0.5 kg of nitrogen are mixed at 20°C and at a total pressure of 100 kPa. Find (a) the volume of the mixture, (b) the partial volumes of
the components, (c) the partial pressures of the components, (d) the mole fractions of the components, (e) the specific heats cp and cv of the mixture, and (f )
the gas constant of the mixture.
10.20 A gaseous mixture consists of 1 kg of oxygen and 2 kg of nitrogen at a pressure
of 150 kPa and a temperature of 20°C. Determine the changes in internal
Properties of Gases and Gas Mixtures
377
energy, enthalpy and entropy of the mixture when the mixture is heated to a
temperature of 100°C (a) at constant volume, and (b) at constant pressure.
10.21 A closed rigid cylinder is divided by a diaphragm into two equal compartments,
each of volume 0.1 m3. Each compartment contains air at a temperature of
20∞C. The pressure in one compartment is 2.5 MPa and in the other compartment is 1 MPa. The diaphragm is ruptured so that the air in both the compartments mixes to bring the pressure to a uniform value throughout the cylinder
which is insulated. Find the net change of entropy for the mixing process.
10.22 A vessel is divided into three compartments (a), (b), and (c) by two partitions.
Part (a) contains oxygen and has a volume of 0.1 m3, (b) has a volume of 0.2 m3
and contains nitrogen, while (c) is 0.05 m3 and holds CO2. All three parts are at
a pressure of 2 bar and a temperature of 13°C. When the partitions are removed
and the gases mix, determine the change of entropy of each constituent, the
final pressure in the vessel and the partial pressure of each gas. The vessel may
be taken as being completely isolated from its surroundings.
Ans. 0.0875, 0.0783, 0.0680 kJ/K; 2 bar; 0.5714, 1.1429, 0.2857 bar.
10.23 A Carnot cycle uses 1 kg of air as the working fluid. The maximum and
minimum temperatures of the cycle are 600 K and 300 K. The maximum pressure of the cycle is 1 MPa and the volume of the gas doubles during the isothermal heating process. Show by calculation of net work and heat supplied that
the efficiency is the maximum possible for the given maximum and minimum
temperatures.
10.24 An ideal gas cycle consists of three reversible processes in the following
sequence: (a) constant volume pressure rise, (b) isentropic expansion to r times
the initial volume, and (c) constant pressure decrease in volume. Sketch the
cycle on the p-v and T-s diagrams. Show that the efficiency of the cycle is
hcycle =
r g - 1 - g (r - 1)
rg - 1
Evaluate the cycle efficiency when g =
4
and r = 8.
3
Ans. (h = 0.378)
10.25 Using the Dieterici equation of state
p=
F
GH
RT
a
◊ exp v-b
RTv
I
JK
(a) Show that
a
a
, vc = 2b, Tc =
2 2
4
Rb
4e b
(b) expand in the form
B C
pv = RT 1 + + 2 +...
v v
(c) show that
a
TB =
bR
10.26 The number of moles, the pressures, and the temperatures of gases a, b, and c
are given below
Gas
m (kg mol)
p (kPa)
t ( °C)
N2
1
350
100
pc =
FG
H
IJ
K
378
Engineering Thermodynamics
CO
3
420
200
O2
2
700
300
If the containers are connected, allowing the gases to mix freely, find (a) the
pressure and temperature of the resulting mixture at equilibrium, and (b) the
change of entropy of each constituent and that of the mixture.
10.27 Calculate the volume of 2.5 kg moles of steam at 236.4 atm. and 776.76 K with
the help of compressibility factor versus reduced pressure graph. At this volume and the given pressure, what would the temperature be in K, if steam behaved like a van der Waals gas?
The critical pressure, volume, and temperature of steam are 218.2 atm.,
57 cm3/g mole, and 647.3 K respectively.
10.28 Two vessels, A and B, each of volume 3 m3 may be connected together by a
tube of negligible volume. Vessel A contains air at 7 bar, 95°C while B contains
air at 3.5 bar, 205°C. Find the change of entropy when A is connected to B.
Assume the mixing to be complete and adiabatic.
Ans. (0.975 kJ/kg K)
10.29 An ideal gas at temperature T1 is heated at constant pressure to T2 and then
expanded reversibly, according to the law pvn = constant, until the temperature
is once again T1 . What is the required value of n, if the changes of entropy
during the separate processes are equal?
2g
Ans. n =
g +1
FG
H
IJ
K
10.30 A certain mass of sulphur dioxide (SO2 ) is contained in a vessel of 0.142 m3
capacity, at a pressure and temperature of 23.1 bar and 18°C respectively. A
valve is opened momentarily and the pressure falls immediately to 6.9 bar.
Sometimes later the temperature is again 18°C and the pressure is observed to
be 9.1 bar. Estimate the value of specific heat ratio.
Ans. 1.29
10.31 A gaseous mixture contains 21% by volume of nitrogen, 50% by volume of
hydrogen, and 29% by volume of carbon-dioxide. Calculate the molecular
weight of the mixture, the characteristic gas constant R for the mixture and the
value of the reversible adiabatic index g. (At 10°C, the cp values of nitrogen,
hydrogen, and carbon dioxide are 1.039, 14.235, and 0.828 kJ/kg K respectively.)
A cylinder contains 0.085 m3 of the mixture at 1 bar and 10°C. The gas undergoes a reversible non-flow process during which its volume is reduced to
one-fifth of its original value. If the law of compression is pv1.2 = constant,
determine the work and heat transfer in magnitude and sense and the change in
entropy.
Ans. 19.64 kg/kg mol, 0.423 kJ/kg K, 1.365,
– 16 kJ, – 7.24 kJ, – 0.31 kJ/kg K
10.32 Two moles of an ideal gas at temperature T and pressure p are contained in a
compartment. In an adjacent compartment is one mole of an ideal gas at temperature 2T and pressure p. The gases mix adiabatically but do not react chemically when a partition separating the compartments is withdrawn. Show that
the entropy increase due to the mixing process is given by
FG 27 + g ln 32 IJ
H 4 g - 1 27 K
R ln
Properties of Gases and Gas Mixtures
379
provided that the gases are different and that the ratio of specific heat g is the
same for both gases and remains constant.
What would the entropy change be if the mixing gases were of the same
species?
10.33 n1 moles of an ideal gas at pressure p1 and temperature T are in one compartment of an insulated container. In an adjoining compartment, separated by a
partition, are n 2 moles of an ideal gas at pressure p2 and temperature T. When
the partition is removed, calculate (a) the final pressure of the mixture, (b) the
entropy change when the gases are identical, and (c) the entropy change when
the gases are different. Prove that the entropy change in (c) is the same as that
produced by two independent free expansions.
10.34 Assume that 20 kg of steam are required at a pressure of 600 bar and a temperature of 750°C in order to conduct a particular experiment. A 140-litre heavy
duty tank is available for storage.
Predict if this is an adequate storage capacity using:
(a) the ideal gas theory,
(b) the compressibility factor chart,
(c) the van der Waals equation with a = 5.454 (litre)2 atm/(g mol)2 , b = 0.03042
litres/gmol for steam,
(d) the Mollier chart
(e) the steam tables.
Estimate the error in each.
10.35 Estimate the pressure of 5 kg of CO2 gas which occupies a volume of 0.70 m3
at 75°C, using the Beattie-Bridgeman equation of state.
Compare this result with the value obtained using the generalized compressibility chart. Which is more accurate and why?
For CO2 with units of atm, litres/g mol and K, A0 = 5.0065, a = 0.07132, B0 =
0.10476, b = 0.07235, C ¥ 10–4 = 66.0.
10.36 Measurements of pressure and temperature at various stages in an adiabatic air
turbine show that the states of air lie on the line pv1.25 = constant. If kinetic and
gravitational potential energy are neglected, prove that the shaft work per kg as
a function of pressure is given by the following relation
LM F p I OP
MN GH p JK PQ
1/ 5
W = 3.5 p1v1 1 -
2
1
Take g for air as 1.4.
10.37 A mass of an ideal gas exists initially at a pressure of 200 kPa, temperature
300 K, and specific volume 0.5 m3/kg. The value of g is 1.4. (a) Determine the
specific heats of the gas. (b) What is the change in entropy when the gas is
expanded to pressure 100 kPa according to the law pv1.3 = const? (c) What will
be the entropy change if the path is pv1.5 = const. (by the application of a cooling jacket during the process)? (d) What is the inference you can draw from
this example?
Ans. (a) 1.166, 0.833 kJ/kg K, (b) 0.044 kJ/kg K (c) – 0.039 kJ/kg K
(d) Entropy increases when n < g and decreases when n > g
10.38 (a) A closed system of 2 kg of air initially at pressure 5 atm and temperature
227°C, expands reversibly to pressure 2 atm following the law pv 1.25 =
380
Engineering Thermodynamics
const. Assuming air as an ideal gas, determine the work done and the heat
transferred.
Ans. 193 kJ, 72 kJ
(b) If the system does the same expansion in a steady flow process,what is the
work done by the system?
Ans. 241 kJ
10.39 Air contained in a cylinder fitted with a piston is compressed reversibly according to the law pv 1.25 = const. The mass of air in the cylinder is 0.1 kg. The initial
pressure is 100 kPa and the initial temperature 20°C. The final volume is 1/8 of
the initial volume. Determine the work and the heat transfer.
Ans. – 22.9 kJ, – 8.6 kJ
10.40 Air is contained in a cylinder fitted with a frictionless piston. Initially the cylinder contains 0.5 m3 of air at 1.5 bar, 20°C. The air is then compressed reversibly according to the law pvn = constant until the final pressure is 6 bar, at
which point the temperature is 120°C. Determine: (a) the polytropic index n,
(b) the final volume of air, (c) the work done on the air and the heat transfer,
and (d) the net change in entropy.
Ans. (a) 1,2685, (b) 0.1676 m3 (c) – 95.3 kJ, – 31.5 kJ, (d) 0.0153 kJ/K
10.41 The specific heat at constant pressure for air is given by
cp = 0.9169 + 2.577 + 10 –4 T – 3.974 ¥ 10–8 T2 kJ/kg K
Determine the change in internal energy and that in entropy of air when it undergoes a change of state from 1 atm and 298 K to a temperature of 2000 K at
the same pressure.
Ans. 1470.4 kJ/kg, 2.1065 kJ/kg K
10.42 A closed system allows nitrogen to expand reversibly from a volume of
0.25 m3 to 0.75 m3 along the path pv 1.32 = const. The original pressure of the
gas is 250 kPa and its initial temperature is 100°C. (a) Draw the p-v and
T-s diagrams.
(b) What are the final temperature and the final pressure of the gas? (c) How
much work is done and how much heat is transferred? (d) What is the
entropy change of nitrogen?
Ans. (b) 262.44 K, 58.63 kPa, (c) 57.89 kJ, 11.4 kJ, (d) 0.0362 kJ/K
10.43 Methane has a specific heat at constant pressure given by cp = 17.66 +
0.06188 T kJ/kg mol K when 1 kg of methane is heated at constant volume
from 27 to 500°C. If the initial pressure of the gas is 1 atm, calculate the final
pressure, the heat transfer, the work done and the change in entropy.
Ans. 2.577 atm, 1258.5 kJ/kg, 0, 2.3838 kJ/kg K
10.44 Air is compressed reversibly according to the law pv1.25 = const. from an initial
pressure of 1 bar and volume of 0.9 m3 to a final volume of 0.6 m3 . Determine
the final pressure and the change of entropy per kg of air.
Ans. 1.66 bar, – 0.0436 kJ/kg K
10.45 In a heat engine cycle, air is isothermally compressed. Heat is then added at
constant pressure, after which the air expands isentropically to its original state.
Draw the cycle on p-v and T-s coordinates. Show that the cycle efficiency can
be expressed in the following form
(g - 1)ln r
h=1–
g [ r g -1/ g - 1]
where r is the pressure ratio, p2/p1 . Determine the pressure ratio and the cycle
efficiency if the initial temperature is 27°C and the maximum temperature is
327°C.
Ans. 13.4, 32.4%
Properties of Gases and Gas Mixtures
381
10.46 What is the minimum amount of work required to separate 1 mole of air at
27°C and 1 atm pressure (assumed composed of 1/5 O2 and 4/5 N2) into oxygen
and nitrogen each at 27°C and 1 atm pressure?
Ans. 1250 J
10.47 A closed adiabatic cylinder of volume 1 m3 is divided by a partition into two
compartments 1 and 2. Compartment 1 has a volume of 0.6 m3 and contains
methane at 0.4 MPa, 40°C, while compartment 2 has a volume of 0.4 m3 and
contains propane at 0.4 MPa, 40°C. The partition is removed and the gases are
allowed to mix. (a) When the equilibrium state is reached, find the entropy
change of the universe. (b) What are the molecular weight and the specific heat
ratio of the mixture?
The mixture is now compressed reversibly and adiabatically to 1.2 MPa. Compute (c) the final temperature of the mixture, (d) the work required per unit
mass, and (e) the specific entropy change for each gas. Take cp of methane and
propane as 35.72 and 74.56 kJ/kg mol K respectively.
Ans. (a) 0.8609 kJ/K, (b) 27.2, 1.193 (c) 100.9°C, (d) 396 kJ, (e) 0.255 kJ/kg K
10.48 An ideal gas cycle consists of the following reversible processes: (i) isentropic
compression, (ii) constant volume heat addition, (iii) isentropic expansion, and
(iv) constant pressure heat rejection. Show that the efficiency of this cycle is
given by
h=1–
LM g (a - 1) OP
N a -1 Q
1/ g
1
rkg where rk is the compression ratio and a is the ratio of pressures after and before
heat addition.
An engine operating on the above cycle with a compression ratio of 6 starts the
compression with air at 1 bar, 300 K. If the ratio of pressures after and before
heat addition is 2.5, calculate the efficiency and the m.e.p. of the cycle. Take
g = 1.4 and cv = 0.718 kJ/kg K.
Ans. 0.579, 2.5322 bar
10.49 The relation between u, p and v for many gases is of the form u = a + bpv where
a and b are constants. Show that for a reversible adiabatic process pv g = constant, where g = (b + 1)/b.
10.50 (a) Show that the slope of a reversible adiabatic process on p-v coordinates is
dp
1 cp
1 ∂v
=where k = dv
kv cv
v ∂p T
(b) Hence, show that for an ideal gas, pv g = constant, for a reversible adiabatic
process.
10.51 A certain gas obeys the Clausius equation of state p (v – b) = RT and has its
internal energy given by u = cv T. Show that the equation for a reversible adiabatic process is p (v – b)g = constant, where g = cp/c v.
10.52 (a) Two curves, one representing a reversible adiabatic process undergone by
an ideal gas and the other an isothermal process by the same gas, intersect
at the same point on the p-v diagram. Show that the ratio of the slope of
the adiabatic curve to the slope of the isothermal curve is equal to g. (b)
Determine the ratio of work done during a reversible adiabatic process to
the work done during an isothermal process for a gas having g = 1.6. Both
processes have a pressure ratio of 6.
1
FG IJ
H K
382
Engineering Thermodynamics
10.53 Two containers p and q with rigid walls contain two different monatomic gases
with masses mp and mq, gas constants Rp and Rq, and initial temperatures Tp and
Tq respectively, are brought in contact with each other and allowed to exchange
energy until equilibrium is achieved. Determine: (a) the final temperature of
the two gases and (b) the change of entropy due to this energy exchange.
10.54 The pressure of a certain gas (photon gas) is a function of temperature only and
is related to the energy and volume by p(T ) = (1/3) (U/V). A system consisting
of this gas confined by a cylinder and a piston undergoes a Carnot cycle between two pressures p1 and p2. (a) Find expressions for work and heat of reversible isothermal and adiabatic processes. (b) Plot the Carnot cycle on p-v and
T-s diagrams. (c) Determine the efficiency of the cycle in terms of pressures.
(d) What is the functional relation between pressure and temperature?
10.55 The gravimetric analysis of dry air is approximately: oxygen = 23%, nitrogen =
77%. Calculate: (a) the volumetric analysis, (b) the gas constant, (c) the molecular weight, (d) the respective partial pressures, (e) the specific volume at
1 atm, 15°C, and (f) How much oxygen must be added to 2.3 kg air to produce
a mixture which is 50% oxygen by volume?
Ans. (a) 21% O2 , 79% N2, (b) 0.288 kJ/kg K,
(d) 21 kPa for O2 , (e) 0.84 m3/kg, (f ) 1.47 kg
10.56 A vessel of volume 2V is divided into two equal compartments. These are filled
with the same ideal gas, the temperature and pressure on one side of the partition being ( p1, T1 ) and on the other ( p2, T2 ). Show that if the gases on the two
sides are allowed to mix slowly with no heat entering, the final pressure and
temperature will be given by
p + p2
T T ( p + p2 )
p= 1
,T= 1 2 1
p1T2 + p2 T1
2
Further, show that the entropy gain is
DS = V
LMF c I R p ln T + p ln T U - p ln p - p ln p OP
MNGH R JK ST T T T T VW T p T p PQ
p
1
1
2
1
2
1
2
1
2
1
2
2
10.57 An ideal gas with a constant volume of cp = 29.6 J/gmol-K is made to undergo
a cycle consisting of the following reversible processes in a closed system:
Process 1–2: The gas expands adiabatically from 5 MPa, 550 K to 1 MPa;
Process 2–3: The gas is heated at constant volume until 550 K; Process 3–1:
The gas is compressed isothermally back to its initial condition.
Calculate the work, the heat and the change of entropy of the gas for each of the
three processes. Draw the p-v and T-s diagrams.
Ans. W1 – 2 = 4260 J/gmol, Q1 – 2 = 0, Ds1 – 2 = 0, W2 – 3 = 0,
Q2 – 3 = 4260 J/gmol, Ds2 – 3 = 9.62 J/g mol-K, W3 – 1 = – 5290 J/mol
= Q3 – 1, Ds3 – 1 = – 9.62 J/gmol-K, Wnet – Qnet = – 1030 g/gmol, f dS = 0.
10.58 Air in a closed system expands reversibly and adiabatically from 3 MPa. 200°C
to two times its initial volume, and then cools at constant volume until the
pressure drops to 0.8 MPa. Calculate the work done and heat transferred per kg
of air. Use cp = 1.017 and cv = 0.728 kJ/kgK. Ans. 82.7 kJ/kg, – 78.1 kJ/kg
10.59 A vessel is divided into three compartments (a), (b) and (c) by two partitions.
Part (a) contains hydrogen and has a volume of 0.1 m3 , part (b) contains nitrogen and has a volume of 0.2 m3 and part (c) contains carbon dioxide and has a
Properties of Gases and Gas Mixtures
383
volume of 0.05 m3. All the three parts are at a pressure of 2 bar and a temperature of 13°C. The partitions are removed and the gases are allowed to mix.
Determine (a) the molecular weight of the mixture, (b) the characteristics gas
constant for the mixture, (c) the partial pressures of each gas, (d) the reversible
adiabatic index g, and (e) the entropy change due to diffusion. The
specific heats of hydrogen, nitrogen and carbon dioxide are 14.235, 1.039 and
0.828 kJ/kg K respectively.
The above gas mixture is then reversibly compressed to a pressure of 6 bar
according to the law pv 1.2 = constant, (f ) Determine the work and heat interactions in magnitude and sense, and (g) the change in entropy.
Ans. (a) 22.8582 (b) 0.3637 kJ/kg K (c) pH2 = 0.5714,
pN2 = 1.1428, pCO2 = 0.2858 bar (d) 1.384 (e) 0.3476 kJ/kgK
(f) – 70.455 kJ, – 33.772 kJ(g) – 0.1063 kJ/K.
10.60 A four cylinder single-stage air compressor has a bore of 200 mm and a stroke
of 300 mm and runs at 400 rpm. At a working pressure of 721.3 kPa it delivers
3.1 m3 of air per min at 270°C. Calculate (a) the mass flow rate, (b) the free air
delivery (FAD) (c) effective swept volume, (d) volumetric efficiency. Take the
inlet condition as that of the free air at 101.3 kPa, 21°C.
Ans. (a) 0.239 kg/s (b) 0.199 m3 /s (c) 0.299 m3 , (d) 79.2%
10.61 Predict the pressure of nitrogen gas at T = 175 K and v = 0.00375 m3 /kg on the
basis of (a) the ideal gas equation of state, (b) the van der Waals equation of
state, (c) the Beattie-Bridgeman equation of state and (d) the Benedict-WebbRubin equation of state. Compare the values obtained with the experimentally
determined value of 10,000 kPa.
Ans. (a) 13,860 kPa (b) 9468 kPa (c) 10,110 kPa (d) 10,000 kPa
10.62 The pressure in an automobile tyre depends on the temperature of the air in the
tyre. When the air temperature is 25°C, the pressure gauge reads 210 kPa. If the
volume of the tyre is 0.025 m3, determine the pressure rise in the tyre when the
air temperature in the tyre rises to 50°C. Also find the amount of air that must
be bled off to restore pressure to its original value at this temperature. Take
atmospheric pressure as 100 kPa.
10.63 Two tanks are connected by a valve. One tank contains 2 kg of CO gas at 77°C
and 0.7 bar. The other tank holds 8 kg of the same gas at 27°C and 1.2 bar. The
valve is opened and the gases are allowed to mix while receiving energy by
heat transfer from the surroundings. The final equilibrium temperature is 42°C.
Using the ideal gas model, determine (a) the final equilibrium pressure, (b) the
heat transfer for the process.
Ans. (a) 1.05 bar (b) 37.25 kJ
11
Thermodynamic Relations,
Equilibrium and Stability
11.1 SOME MATHEMATICAL THEOREMS
Theorem 1. If a relation exists among the variables x, y, and z, then z may be
expressed as a function of x and y, or
FG ∂ z IJ dx + FG ∂z IJ dy
H ∂xK
H ∂ yK
FG ∂z IJ = M, and FG ∂z IJ = N
H ∂yK
H ∂x K
dz =
x
y
If
y
then
x
dz = M dx + N dy,
where z, M and N are functions of x and y. Differentiating M partially with respect
to y, and N with respect to x
FG ∂ M IJ = ∂ z
H ∂y K ∂x ◊ ∂y
FG ∂ N IJ = ∂ z
H ∂x K ∂y ◊ ∂x
FG ∂ M IJ = FG ∂ N IJ
H ∂ y K H ∂x K
2
x
2
y
\
x
(11.1)
y
This is the condition of exact (or perfect) differential.
Theorem 2. If a quantity f is a function of x, y, and z, and a relation exists among
x, y and z, then f is a function of any two of x, y, and z. Similarly any one of x, y, and
z may be regarded to be a function of f and any one of x, y, and z. Thus, if
x = x ( f, y)
dx =
FG ∂ x IJ df + FG ∂ x IJ dy
H ∂ f K H ∂yK
y
f
Thermodynamic Relations, Equilibrium and Stability
385
Similarly, if
y = y ( f, z )
dy =
FG ∂ y IJ df + FG ∂ y IJ dz
H ∂ f K H ∂z K
z
f
Substituting the expression of dy in the preceding equation
FG ∂ x IJ df + FG ∂ x IJ LMFG ∂ y IJ df + FG ∂ y IJ dzOP
H ∂ f K H ∂ y K MNH ∂ f K H ∂ z K PQ
LF ∂ x I F ∂ x I F ∂ y I O F ∂ x I F ∂ y I
= MG J + G J G J P df + G J G J dz
MNH ∂ f K H ∂ y K H ∂ f K PQ H ∂ y K H ∂z K
dx =
y
f
y
z
f
f
z
f
f
Again
FG ∂ x IJ df + FG ∂ x IJ dz
H ∂ f K H ∂z K
FG ∂ x IJ = FG ∂ x IJ FG ∂ y IJ
H ∂ z K H ∂ y K H ∂z K
FG ∂ x IJ FG ∂ y IJ FG ∂ z IJ = 1
H ∂ y K H ∂z K H ∂ x K
dx =
z
\
f
\
f
f
f
f
f
(11.2)
f
Theorem 3. Among the variables x, y, and z, any one variable may be considered
as a function of the other two. Thus
x = x ( y, z)
dx =
FG ∂ x IJ dy + FG ∂ x IJ dz
H ∂y K H ∂z K
z
y
Similarly,
FG ∂ z IJ dx + FG ∂z IJ dy
H ∂xK
H ∂ yK
F ∂ x I F ∂ x I LF ∂z I F ∂ z I O
dx = G J dy + G J MG J dx + G J dyP
H ∂ y K H ∂z K MNH ∂ x K H ∂ y K PQ
LF ∂ x I F ∂ x I F ∂z I O F ∂ x I F ∂ z I
= MG J + G J G J P dy + G J G J dx
MNH ∂ y K H ∂z K H ∂ y K PQ H ∂ z K H ∂ x K
LF ∂ x I F ∂ x I F ∂z I O
= MG J + G J G J P dy + dx
MNH ∂ y K H ∂z K H ∂ y K PQ
dz =
\
y
x
z
y
y
z
y
x
z
y
x
x
y
y
386
Engineering Thermodynamics
FG ∂ x IJ + FG ∂ x IJ FG ∂z IJ = 0
H ∂ y K H ∂z K H ∂ y K
FG ∂ x IJ FG ∂ z IJ FG ∂ y IJ = – 1
H ∂ y K H ∂ x K H ∂z K
\
z
or
y
z
y
x
(11.3)
x
Among the thermodynamic variables p, V and T, the following relation holds good
∂p
∂V
∂T
=–1
∂V T ∂T p ∂ p V
FG IJ FG IJ FG IJ
H K H K H K
11.2
MAXWELL’S EQUATIONS
A pure substance existing in a single phase has only two independent variables. Of
the eight quantities p, V, T, S, U, H, F (Helmholtz function), and G (Gibbs function)
any one may be expressed as a function of any two others.
For a pure substance undergoing an infinitesimal reversible process
(a) dU = TdS – pdV
(b) dH = dU + pdV + Vdp = TdS + Vdp
(c) dF = dU – TdS – SdT = – pdV – SdT
(d) dG = dH – TdS – SdT = Vdp – SdT
Since U, H, F and G are thermodynamic properties and exact differentials of the
type
dz = M dx + N dy, then
FG ∂ M IJ = FG ∂ N IJ
H ∂y K H ∂x K
x
y
Applying this to the four equations
FG ∂T IJ = – FG ∂ p IJ
H ∂V K H ∂ S K
FG ∂T IJ = FG ∂V IJ
H ∂ p K H ∂S K
FG ∂ p IJ = FG ∂S IJ
H ∂ T K H ∂V K
FG ∂V IJ = – FG ∂S IJ
H ∂T K H ∂ pK
S
S
p
V
T
p
(11.4)
V
(11.5)
(11.6)
(11.7)
T
These four equations are known as Maxwell’s equations.
11.3 TDS EQUATIONS
Let entropy S be imagined as a function of T and V. Then
387
Thermodynamic Relations, Equilibrium and Stability
FG ∂S IJ dT + FG ∂S IJ dV
H ∂T K
H ∂V K
F ∂S I
F ∂S I
TdS = T G J dT + T G J dV
H ∂T K
H ∂V K
dS =
V
\
T
V
Since T
T
FG ∂S IJ = C , heat capacity at constant volume, and
H ∂T K
FG ∂S IJ = FG ∂ p IJ , Maxwell’s third equation,
H ∂V K H ∂T K
F ∂ pI
TdS = C dT + T G J dV
H ∂T K
v
V
T
V
(11.8)
v
V
This is known as the first TdS equation.
If S = S (T, p)
FG ∂ S IJ dT + FG ∂S IJ dp
H ∂ pK
H ∂T K
F ∂S I
F ∂S I
TdS = T G J dT + T G J dp
H ∂ pK
H ∂T K
F ∂ S I F ∂V I
F ∂S I
T G J = C , and G J = – G J
H ∂T K
H ∂ p K H ∂T K
dS =
T
p
\
p
Since
T
p
T
p
p
then
TdS = Cp dT – T
FG ∂V IJ dp
H ∂T K
(11.9)
p
This is known as the second TdS equation.
11.4 DIFFERENCE IN HEAT CAPACITIES
Equating the first and second TdS equations
FG ∂V IJ dp = C dT + T FG ∂ p IJ dV
H ∂T K
H ∂T K
F ∂ pI
F ∂V I
(C – C ) dT = T G J dV + T G J dp
H ∂T K
H ∂T K
F ∂V IJ
F ∂pI
TG
TG J
H ∂T K dV + H ∂T K dp
dT =
TdS = Cp dT – T
v
V
p
p
v
V
\
p
V
p
C p - Cv
C p - Cv
388
Again
Engineering Thermodynamics
FG ∂T IJ dV + FG ∂T IJ dp
H ∂V K
H ∂pK
F ∂V IJ
F ∂pI
TG
TG J
H ∂T K = F ∂T I and H ∂T K = FG ∂T IJ
GH ∂V JK
H ∂ pK
C -C
C -C
dT =
p
\
v
p
V
p
v
p
p
v
V
Both these equations give
FG ∂ p IJ FG ∂V IJ
H ∂T K H ∂ T K
FG ∂ p IJ FG ∂T IJ FG ∂V IJ = – 1
H ∂ T K H ∂V K H ∂ p K
F ∂V I F ∂ p I
C –C =–TG J G J
H ∂ T K H ∂V K
Cp – C v = T
p
v
But
p
v
T
2
\
p
(11.10)
v
T
T
This is a very important equation in thermodynamics. It indicates the following
important facts.
2
(a) Since
FG ∂V IJ is always positive, and FG ∂ p IJ for any substance is negative,
H ∂T K
H ∂V K
p
T
(Cp – Cv) is always positive. Therefore, Cp is always greater than Cv.
(b) As T Æ 0 K, Cp Æ Cv or at absolute zero, Cp = Cv .
(c) When
FG ∂V IJ = 0 (e.g., for water at 4°C, when density is maximum, or speH ∂T K
p
cific volume minimum), Cp = Cv.
(d) For an ideal gas, pV = mRT
FG ∂V IJ = m R = V
H ∂T K p T
FG ∂ p IJ = – mRT
V
H ∂V K
p
and
2
T
\
Cp – Cv = mR
or
cp – cv = R
Equation (11.10) may also be expressed in terms of volume expansivity (b),
defined as
b=
FG IJ
H K
1 ∂V
V ∂T
p
Thermodynamic Relations, Equilibrium and Stability
389
and isothermal compressibility (kT ), defined as
1 ∂V
FG IJ
V H ∂ pK
L 1 F ∂V I O
TV M G J P
MNV H ∂T K PQ
C –C =
1 F ∂V I
- G J
V H ∂ pK
kT = –
T
2
p
p
v
T
\
Cp – C v =
TVb
2
(11.11)
kT
11.5 RATIO OF HEAT CAPACITIES
At constant S, the two TdS equations become
FG ∂V IJ dp
H ∂T K
F ∂pI
C dT = – T G J dV
H ∂T K
Cp dTs = T
s
p
s
v
s
V
\
FG ∂ p IJ
C
F ∂V IJ FG ∂T IJ FG ∂ p IJ = H ∂V K = g
= –G
H ∂T K H ∂ p K H ∂V K FG ∂ p IJ
C
H ∂V K
p
S
v
p
V
S
T
Since g > 1,
FG ∂ p IJ > FG ∂ p IJ
H ∂V K H ∂V K
S
T
Therefore, the slope of an isentrope is greater than that of an isotherm on p – v
diagram (Fig. 11.1). For reversible and adiabatic compression, the work done is
Ws = h2s – h1 =
z
2S
1
vdp
= Area 1 – 2S – 3 – 4 – 1
For reversible and isothermal compression, the work done would be
WT = h2T – h1 =
z
2T
1
vdp
= Area 1–2T – 3 – 4 – 1
\
WT < WS
390
Engineering Thermodynamics
p
3
p2
2S
2T
S=C
T=C
4
1
p1
v
Fig. 11.1 Compression Work in Different Reversible Process
For polytropic compression with 1 < n < g, the work done will be between these two
values. So, isothermal compression requires minimum work.
The adiabatic compressibility (ks) is defined as
FG IJ
H K
1 F ∂V I
- G J
C
V H ∂p K
=
=g
C
1 F ∂V I
– G J
V H ∂p K
ks = –
\
1 ∂V
V ∂p s
p
T
v
s
or
k
g=
ks
(11.12)
11.6 ENERGY EQUATION
For a system undergoing an infinitesimal reversible process between two equilibrium states, the change of internal energy is
dU = TdS – pdV
Substituting the first TdS equation
FG ∂p IJ dV – pdV
H ∂T K
L F ∂p I O
= C dT + MT G J - p P dV
N H ∂T K Q
dU = Cv dT + T
V
v
(11.13)
V
If
U = U (T, V)
FG ∂U IJ dT + FG ∂U IJ dV
H ∂T K H ∂V K
FG ∂U IJ = T FG ∂p IJ – p
H ∂V K H ∂ T K
dU =
T
V
T
V
(11.14)
Thermodynamic Relations, Equilibrium and Stability
391
This is known as the energy equation. Two applications of the equation are given
below:
nRT
(a) For an ideal gas,
p=
V
∂p
nR p
=
\
=
∂T V
V
T
\
FG IJ
H K
FG ∂U IJ = T ◊ p – p = 0
H ∂V K
T
T
U does not change when V changes at T = C.
FG ∂U IJ FG ∂p IJ FG ∂V IJ = 1
H ∂p K H ∂V K H ∂U K
FG ∂U IJ FG ∂p IJ = FG ∂U IJ = 0
H ∂p K H ∂V K H ∂V K
FG ∂p IJ π 0, FG ∂U IJ = 0
H ∂V K H ∂p K
T
T
\
T
T
Since
T
T
T
T
U does not change either when p changes at T = C. So the internal energy of an ideal
gas is a function of temperature only, as shown earlier in Chapter 10.
Another important point to note is that in Eq. (11.13), for an ideal gas
pV = nRT and T
FG ∂p IJ – p = 0
H ∂T K
v
Therefore
dU = Cv dT
holds good for an ideal gas in any process (even when the volume changes). But for
any other substance
dU = Cv dT
is true only when the volume is constant and dV = 0.
Similarly
dH = TdS + Vdp
and
FG ∂V IJ dp
H ∂T K
L F ∂V I O
dH = C dT + MV - T G J P dp
MN H ∂T K PQ
FG ∂H IJ = V – T FG ∂V IJ
H ∂T K
H ∂p K
TdS = Cp dT – T
p
\
p
(11.15)
p
\
T
(11.16)
p
As shown for internal energy, it can be similarly proved from Eq. (11.16) that the
enthalpy of an ideal gas is not a function of either volume or pressure
392
Engineering Thermodynamics
LMi.e. F ∂H I = 0 and FG ∂H IJ = 0OP
H ∂V K PQ
MN GH ∂p JK
T
T
but a function of temperature alone.
Since for an ideal gas, pV = nR T
and
V–T
FG ∂V IJ = 0
H ∂T K
p
the relation dH = Cp dT is true for any process (even when the pressure changes).
However, for any other substance the relation dH = Cp dT holds good only when the
pressure remains constant or dp = 0.
(b) Thermal radiation in equilibrium with the enclosing walls possesses an energy that depends only on the volume and temperature. The energy density (u),
defined as the ratio of energy to volume, is a function of temperature only, or
U
u=
= f (T) only.
V
The electromagnetic theory of radiation states that radiation is equivalent to a photon gas and it exerts a pressure, and that the pressure exerted by the black-body
radiation in an enclosure is given by
u
p=
3
Black-body radiation is thus specified by the pressure, volume, and temperature of
the radiation.
u
Since
U = uV and p =
3
∂U
∂p
1 du
= u and
=
∂V T
∂T V 3 dT
By substituting in the energy Eq. (11.13)
T du u
u=
3 dT 3
FG IJ
H K
FG IJ
H K
\
du
dT
=4
u
T
or
ln u = ln T 4 + ln b
or
u = bT 4
where b is a constant. This is known as the Stefan-Boltzmann Law.
Since
U = uV = VbT 4
FG ∂U IJ = C = 4VbT
H ∂T K
FG ∂p IJ = 1 du = 4 bT
H ∂T K 3 dT 3
3
v
V
and
V
3
Thermodynamic Relations, Equilibrium and Stability
393
from the first TdS equation
TdS = Cv dT + T
FG ∂p IJ dV
H ∂T K
V
4
= 4VbT 3 dT + bT 4 ◊ dV
3
For a reversible isothermal change of volume, the heat to be supplied reversibly to
keep temperature constant
4
Q = bT 4 DV
3
For a reversible adiabatic change of volume
4
bT 4 dV = – 4VbT 3 dT
3
dV
dT
\
= –3
V
T
3
or
VT = const.
If the temperature is one-half the original temperature, the volume of black-body
radiation is to be increased adiabatically eight times its original volume so that the
radiation remains in equilibrium with matter at that temperature.
11.7 JOULE-KELVIN EFFECT
A gas is made to undergo continuous throttling process by a valve, as shown in
Fig. 11.2. The pressures and temperatures of the gas in the insulated pipe upstream
and downstream of the valve are measured with suitable manometers and thermometers.
Valve
Let pi and Ti be the arbitrarily chosen
ti
tf
pressure and temperature before throttling
and let them be kept constant. By operating the valve manually, the gas is throttled
pi
pf
successively to different pressures and
temperatures pf 1, Tf 1; pf 2, Tf 2; pf 3, Tf 3 and
Insulation
so on. These are then plotted on the T-p
coordinates as shown in Fig. 11.3. All the
Fig. 11.2 Joule-Thomson Expansion
points represent equilibrium states of
some constant mass of gas, say, 1 kg, at which the gas has the same enthalpy.
The curve passing through all these points is an isenthalpic curve or an isenthalpe.
It is not the graph of a throttling process, but the graph through points of equal
enthalpy.
The initial temperature and pressure of the gas (before throttling) are then set to
new values, and by throttling to different states, a family of isenthalpes is obtained
for the gas, as shown in Figs. 11.4 and 11.5. The curve passing through the maxima
of these isenthalpes is called the inversion curve.
394
Engineering Thermodynamics
States after throttling
f4
f3
T
f5
f2
f6
f1
State before
throttling
f7
h=C
p
Fig. 11.3 Isenthalpic States of a Gas
The numerical value of the slope of an isenthalpe on a T-p diagram at any point
is called the Joule-Kelvin coefficient and is denoted by mJ. Thus the locus of all
points at which mJ is zero is the inversion curve. The region inside the inversion
curve where mJ is positive is called the cooling region and the region outside where
mJ is negative is called the heating region. So,
mJ =
FG ∂T IJ
H ∂p K
h
Maximum
Inversion
Temp
Constant enthalpy
curves (isenthalpes)
T
T1
Cooling
region
+ mJ
Heating region
Critical
point
p
va
Liq
– mJ
T2
Inversion curve (m = 0)
p
p
Fig. 11.4 Isenthalpic Curves and the Inversion Curve
The difference in enthalpy between two neighbouring equilibrium states is
dh = Tds + vdp
and the second TdS equation (per unit mass)
FG ∂v IJ dp
H ∂T K
L F ∂v I O
dh = c dT – MT G J - v P dp
MN H ∂T K PQ
Tds = cp dT – T
p
\
p
p
Thermodynamic Relations, Equilibrium and Stability
395
T
Inversion curve ( m J = 0)
Constant enthalpy
curves
Critical point
Saturation curve
Liquid Vapour
region
s
Fig. 11.5 Inversion and Saturation Curves on T-s Plot
The second term in the above equation stands only for a real gas, because for an
ideal gas, dh = cp dT.
mJ =
FG ∂T IJ = 1 LMT FG ∂v IJ - vOP
H ∂p K c MN H ∂T K PQ
h
p
(11.17)
p
For an ideal gas
pv = RT
\
FG ∂v IJ = R = v
H ∂T K p T
1 F v
T ◊ - vI = 0
m =
H
K
c
T
p
\
J
p
There is no change in temperature when an ideal gas is made to undergo a JouleKelvin expansion (i.e. throttling).
For achieving the effect of cooling by Joule-Kelvin expansion, the initial temperature of the gas must be below the point where the inversion curve intersects the
temperature axis, i.e. below the maximum inversion temperature. For nearly all
substances, the maximum inversion temperature is above the normal ambient
temperature, and hence cooling can be obtained by the Joule-Kelvin effect. In the
case of hydrogen and helium, however, the gas is to be precooled in heat exchangers below the maximum inversion temperature before it is throttled. For liquefaction, the gas has to be cooled below the critical temperature.
Let the initial state of gas before throttling be at A (Fig. 11.6). The change in
temperature may be positive, zero, or negative, depending upon the final pressure
after throttling. If the final pressure lies between A and B, there will be a rise in
temperature or heating effect. If it is at C, there will be no change in temperature. If
the final pressure is below pC, there will be a cooling effect, and if the final pressure
is pD, the temperature drop will be (TA – TD).
Maximum temperature drop will occur if the initial state lies on the inversion
curve. In Fig. 11.6, it is (TB – TD).
396
Engineering Thermodynamics
Cooling
region
Inversion curve
T
B
A
C
Isenthalpe
D
Heating
region
pD pC
pB
pA
p
Fig. 11.6
Maximum Cooling by Joule-Kelvin Expansion
The volume expansivity is
b=
1 ∂v
v ∂T
FG IJ
H K
p
So the Joule-Kelvin coefficient mJ is given by, from Equation (11.17)
mJ =
or
mJ =
LM FG IJ - vOP
MN H K PQ
∂v
1
T
∂T
cp
p
v
[bT – 1]
cp
1
and mJ = 0
T
There are two inversion temperatures for each pressure, e.g. T1 and T2 at pressure p
(Fig. 11.4).
For an ideal gas,
b=
11.8 CLAUSIUS-CLAPEYRON EQUATION
During phase transitions like melting, vaporization and sublimation, the temperature and pressure remain constant, while the entropy and volume change. If x is the
fraction of initial phase i which has been transformed into final phase f, then
s = (1 – x) s (i) + xs( f )
v = (1 – x) v (i) + xv (f )
where s and v are linear functions of x.
For reversible phase transition, the heat transferred per mole (or per kg) is the
latent heat, given by
l = T{s (f ) – s (i)}
which indicates the change in entropy.
Now
dg = – sdT + vdp
or
s= -
FG ∂g IJ
H ∂T K
p
Thermodynamic Relations, Equilibrium and Stability
and
v=
FG ∂g IJ
H ∂p K
397
T
A phase change of the first order is known as any phase change that satisfies the
following requirements:
(a) There are changes of entropy and volume.
(b) The first-order derivatives of Gibbs function change discontinuously.
Let us consider the first-order phase transition of one mole of a substance from
phase i to phase f. Using the first TdS equation
Tds = cv dT + T
FG ∂p IJ dv
H ∂T K
V
for the phase transition which is reversible, isothermal and isobaric, and integrating
over the whole change of phase, and since
FG ∂p IJ is independent of v
H ∂T K
V
T {s ( f) – s (i)} = T
dp
{v (f) – v (i)}
dT
l
dp
s( f ) - s(i )
= (f)
=
(11.18)
(f)
(i)
dT
v -v
T{v - v ( i )}
The above equation is known as the Clausius-Clapeyron equation.
The Clausius-Clapeyron equation can also be derived in another way. For a reversible process at constant T and p, the Gibbs function remains constant. Therefore, for the first-order phase change at T and p
g (i) = g (f)
and for a phase change at T + dT and p + dp (Fig. 11.7)
g (i) + dg (i) = g (f) + dg (f)
p
\
p + dp
Critical point
L
p
V
S
Tr.pt.
T
Fig. 11.7
T + dT
T
First Order Phase Transition
Subtracting
dg (i) = dg (f)
or
– s (i) dT + v (i) dp
= – s (f) dT + v (f) dp
398
Engineering Thermodynamics
dp
s( f ) - s(i )
l
= (f)
=
dT
v - v(i)
T (v( f ) - v(i ) )
\
For fusion
l fu
dp
=
dT
T ( v¢¢ - v ¢ )
where lfu is the latent heat of fusion, the first prime in v, i.e., v¢ indicates the saturated solid state, and the second prime the saturated liquid state. The slope of the
fusion curve is determined by (v¢¢ – v¢), since lfu and T are positive. If the substance
expands on melting, v¢¢ > v¢, the slope is positive. This is the usual case. Water,
however, contracts on melting and has the fusion curve with a negative slope
(Fig. 11.8).
Fusion curve
For any other substance
(positive slope)
For water
Critical point
(negative
slope)
p
Solid
Liquid
p1
Vaporization curve
dp
dT
Sublimation
curve
Vapour
Triple point
T
T1
Fig. 11.8 Phase Diagram for Water and any Other Substance on p-T Coordinates
For vaporization
lvap
dp
=
dT
T ( v ¢¢¢ - v ¢¢ )
where lvap is the latent heat of vaporization, and the third prime indicates the saturated vapour state.
dp
\
lvap = T
(v¢¢¢ – v¢¢)
dT
At temperature considerably below the critical temperature, v¢¢¢ >> v¢¢ and using
the ideal gas equation of state for vapour
RT
v¢¢¢ =
p
dp RT
l vap @ T ◊
dT p
or
lvap =
RT 2 dp
p dT
(11.19)
Thermodynamic Relations, Equilibrium and Stability
399
If the slope dp/dT at any state (e.g. point p1, T1 in Fig. 11.8) is known, the latent
heat of vaporization can be computed from the above equation.
The vapour pressure curve is of the form
B
ln p = A +
+ C ln T + DT
T
where A, B, C and D are constants. By differentiating with respect to T
B C
1 dp
= - 2 + +D
(11.20)
T
T
p dT
Equations (11.19) and (11.20) can be used to estimate the latent heat of vaporization.
Clapeyron’s equation can also be used to estimate approximately the vapour
pressure of a liquid at any arbitrary temperature in conjunction with a relation for
the latent heat of a substance, known as Trouton’s rule, which states that
h fg
@ 88 kJ/kg mol K
TB
where hfg is the latent heat of vaporization in kJ/kg mol and TB is the boiling point
at 1.013 bar. On substituting this into Eq. (11.19)
88TB
dp
=
p
dT
RT2
p
dp 88TB T dT
or
=
101. 325 p
R TB T 2
z
ln
z
FG
H
88TB 1 1
p
= 101.325
R T TB
IJ
K
È 88 Ê T ˆ ˘
p = 101.325 exp Í Á1 - B ˜ ˙
(11.21)
T ¯˚
ÎR Ë
This gives the vapour pressure p in kPa at any temperature T.
For sublimation
dp
lsub
=
dT
T ( v ¢¢¢ - v ¢ )
where lsub is the latent heat of sublimation.
RT
Since v¢¢¢ >> v¢, and vapour pressure is low, v¢¢¢ =
p
dp
lsub
=
RT
dT
T◊
p
d (log p)
or
l sub = – 2.303 R
d (1/ T )
the slope of log p vs. 1/T curve is negative, and if it is known, lsub can be estimated.
At the triple point (Fig. 9.12),
lsub = lvap + lfus
(11.23)
\
F dp I
H dT K
=
vap
ptplvap
RTtp2
400
Engineering Thermodynamics
F dp I
H dT K
=
sub
ptplsub
RTtp2
Since lsub > lvap , at the triple point
dp
dp
>
dT sub
dT vap
Therefore, the slope of the sublimation curve at the triple point is greater than that
of the vaporization curve (Fig. 11.8).
F I
H K
F I
H K
11.9 EVALUATION OF THERMODYNAMIC PROPERTIES
FROM AN EQUATION OF STATE
Apart from calculating pressure, volume, or temperature, an equation of state can
also be used to evaluate other thermodynamic properties such as internal energy,
enthalpy and entropy. The property relations to be used are:
LM FG ∂p IJ - pOP dv
N H ∂T K Q
L F ∂v I O
dh = c dT + Mv - T G J P dp
MN H ∂T K PQ
1 L
F ∂p I O
ds = Mc dT + T G J dv P
H ∂T K Q
T N
1 L
F ∂v I O
= Mcp dT - T G J dpP
H ∂T K QP
T NM
du = cv dT + T
(11.24)
v
(11.25)
p
p
v
v
(11.26)
p
Integrations of the differential relations of the properties p, v and T in the above
equations are carried out with the help of an equation of state. The changes in properties are independent of the path and depend only on the end states. Let us consider that the change in enthalpy per unit mass of a gas from a reference state 0 at
p0, T0 having enthalpy, h0 to some other state B at p, T with enthalpy h is to be
calculated (Fig. 11.9). The reversible path 0B may be replaced for convenience by
either path 0-a-B or path 0-b-B, both also being reversible.
Path 0-a-B:
From Eq. 11.25,
T
ha – h0 =
z
z LMMN FGH IJK OPPQ
c p dT
T0
p
h – ha =
p0
v-T
∂v
∂T
dp
p T
On addition
h – h0 =
LM c dT OP + R|S LMv - T FG ∂v IJ OP dpU|V
N
Q T| MN H ∂T K PQ W|
z
T
T0
p
p0
z
p
p0
p
T
(11.27)
Thermodynamic Relations, Equilibrium and Stability
b
p
(p,T0)
401
p=C
B (p,T )
T0 = C
T=C
p0 = C
O
a( p0,T )
( p0,T0 )
T
Fig. 11.9 Processes Connecting States (p0, T0 ) and (p, T)
Similarly, for
Path 0-b-B:
h – h0 =
R| Lv - T F ∂v I O dpU| + L c dT O
S| MM GH ∂T JK PP V| MN
PQ
Q W
T N
z
p
p0
z
T
T0
p
(11.28)
p
T0
p
Equation (11.27) is preferred to Eq. (11.28) since cp at lower pressure can be
conveniently measured. Now,
z
pv
d (pv) =
p0 va
z
z
v
p dv +
va
z
p0
v
va
p
v dp = pv – p0va –
v dp
LM pdvOP
N Q
z
v
va
(11.29)
T
Again,
LM ∂v OP LM ∂T OP LM ∂p OP = – 1
N ∂T Q N ∂p Q N ∂v Q
LM ∂v OP = - LM ∂p OP LM ∂v OP
N ∂T Q N ∂T Q N ∂p Q
p
\
T
v
v
p
T
Substituting in Eq. (11.27),
h – h0 =
LM c dT OP + pv – p v – LM pdvOP
N Q
N
Q
R| F ∂p I F ∂v I U|
+ S T G J G J dp V
|T H ∂T K H ∂p K |W
z
T
T0
z
0 a
p
va
p0
z
p
p0
p
T
v
T
T
402
Engineering Thermodynamics
=
LM c dT OP + pv – p v – R|S LM p - T FG ∂p IJ OP dvU|V
N
Q
T| N H ∂T K Q W|
z
T
T0
z
0 a
p
v
va
pO
v
(11.30)
T
To find the entropy change, Eq. (11.26) is integrated to yield:
s – s0 =
=
=
LM c dT OP - LM FG ∂v IJ dpOP
N T Q MN H ∂T K PQ
z
LM
Nz
LMz
N
T
T0
cp
T
T0
cp
p
p0
p0
T
T0
z
p
p
T
dT
T
OP - LM- FG ∂p IJ FG ∂v IJ dpOP
Q MN H ∂T K H ∂p K PQ
dT
T
OP + LM FG ∂p IJ dv OP
Q N H ∂T K Q
z
p
p0
p0
z
T
T
v
va
p0
v
v
(11.31)
T
11.10 GENERAL THERMODYNAMIC CONSIDERATIONS ON
AN EQUATION OF STATE
Certain general characteristics are common to all gases. These must be clearly observed in the developing and testing of an equation of state. It is edifying to discuss
briefly some of the more important ones:
(i) Any equation of state must reduce to the ideal gas equation as pressure approaches zero at any temperature. This is clearly seen in a generalized compressibility factor chart in which all isotherms converge to the point z = 1 at zero pres
sure. Therefore,
lim
p Æ0
LM pv OP = 1 at any temperature
N RT Q
Also, as seen from Fig. 10.6, the reduced isotherms approach the line z = 1 as the
temperature approaches infinity, or:
lim
T Æ•
LM pv OP = 1 at any pressure.
N RT Q
(ii) The critical isotherm of an equation of state should have a point of inflection
at the critical point on p–v coordinates, or
LM ∂p OP
N ∂v Q
LM ∂ p OP
N ∂v Q
2
= 0 and
T = Tc
=0
2
T = Tc
(iii) The isochores of an equation of state on a p–T diagram should be essentially
straight, or:
LM ∂p OP = constant, LM ∂ p OP = 0 as p Æ 0, or as T Æ •.
N ∂T Q
N ∂T Q
2
2
v
v
Thermodynamic Relations, Equilibrium and Stability
403
An equation of state can predict the slope of the critical isochores of a fluid. This
slope is identical with the slope of the vaporization curve at the critical point. From
the Clapegron equation, dp/dT = Ds/Dv, the slope of the vaporization curve at the
critical point becomes:
∂s
∂p
dp
=
=
∂v Tc
∂T vc
dT
LM OP
N Q
LM OP
N Q
(by Maxwell’s equation)
Therefore, the vapour-pressure slope at the critical point, dp/dT, is equal to the
slope of the critical isochore (∂p/∂T) vc (Fig. 11.10)
vc
v=
oint
Criti
cal p
v < vc
p
v > vc
T
Fig. 11.10
Pressure-Temperature Diagram with Isochoric Lines
(iv) The slopes of the isotherms of an equation of state on a Z–p compressibility
factor chart as p approaches zero should be negative at lower temperatures and
positive at higher temperatures. At the Boyle temperature, the slope is zero as p
approaches zero, or
lim
pÆ 0
LM ∂z OP = 0 at T = T
N ∂p Q
B
T
An equation of state should predict the Boyle temperature which is about
2.54 Tc for many gases.
An isotherm of maximum slope on the Z–p plot as p approaches zero, called the
foldback isotherm, which is about 5Tc for many gases, should be predicted by an
equation of state, for which:
404
Engineering Thermodynamics
LM ∂ z OP = 0 at T = T
N ∂T ◊ ∂p Q
2
lim
p Æ0
F
where TF is the foldback temperature (Fig. 10.5 a). As temperature increases beyond TF the slope of the isotherm decreases, but always remains positive.
(v) An equation of state should predict the Joule-Thomson coefficient, which is
mJ =
LM FG IJ - vOP = RT FG ∂z IJ
MN H K PQ pc H ∂T K
2
∂v
1
T
∂T
cp
p
p
p
For the inversion curve, mJ = 0,
or,
FG ∂z IJ = 0
H ∂T K
p
11.11
MIXTURES OF VARIABLE COMPOSITION
Let us consider a system containing a mixture of substances 1, 2, 3 ... K. If some
quantities of a substance are added to the system, the energy of the system will
increase. Thus for a system of variable composition, the internal energy depends
not only on S and V, but also on the number of moles (or mass) of various constituents of the system.
Thus
U = U (S, V, n1, n2, ..., nK)
where n1, n2, ..., nK are the number of moles of substances 1, 2, ..., K. The composition may change not only due to addition or subtraction, but also due to chemical
reaction and inter-phase mass transfer. For a small change in U, assuming the function to be continuous.
dU =
FG ∂U IJ
F ∂U IJ
dS + G
H ∂S K
H ∂V K
F ∂U IJ
F ∂U IJ
+G
dn + G
H ∂n K
H ∂n K
F ∂U IJ
+ ... + G
dn
H ∂n K
V , n1 , n2 ,..., n K
dV
S , n1 , n2 ,..., n K
dn2
1
1
2
S , V , n2 ,..., nK
S , V , n1 , n3 ,..., n K
K
K
or
dU =
FG ∂U IJ
H ∂S K
S , V , n1 , n2 ,..., nK -1
FG ∂U IJ dV + Â FG ∂U IJ
H ∂V K
H ∂n K
K
dS +
V , ni
S , ni
i =1
i
dni
S , V , nJ
where subscript i indicates any substance and subscript j any other substance except the one whose number of moles is changing.
If the composition does not change
dU = TdS – pdV
Thermodynamic Relations, Equilibrium and Stability
\
FG ∂U IJ
H ∂S K
= T,
FG ∂U IJ
H ∂V K
and
V , ni
K
\
=–p
S , ni
F ∂U I
 GH ∂n JK
dU = TdS – pdV +
i =1
405
dni
i
(11.32)
S,V , nj
Molal chemical potential, m, of component i is defined as
mi =
FG ∂U IJ
H ∂n K
i
S ,V , n j
signifying the change in internal energy per unit mole of component i when S, V,
and the number of moles of all other components are constant.
K
\
 mi dni
dU = TdS – pdV +
i =1
K
or
 mi dni
TdS = dU + pdV –
(11.33)
i =1
This is known as Gibbs entropy equation.
In a similar manner
G = G (p, T, n1, n2, ..., nK )
FG ∂G IJ dT + Â FG ∂G IJ
H ∂T K
H ∂n K
F ∂G IJ dn
= Vdp – SdT +  G
H ∂n K
or
dG =
FG ∂G IJ
H ∂p K
K
dp +
p , ni
T , ni
i =1
i
dni
T , p, n j
K
i
Since
i =1
i
K
F ∂G I
G = U + pV – TS
d (U + pV – TS) = VdP – SdT +
 GH ∂n JK
i =1
or
T , p, n j
i
dni
T , p, n j
dU + pdV + Vdp – TdS – SdT
K
= Vdp – SdT +
i =1
K
or
dU = TdS – pdV +
Comparing this equation with Eq. (11.32)
i
=
S, V , n j
F ∂G I
GH ∂n JK
i
= mi
T , p, n j
i
dni
T , p, n j
F ∂G I
 GH ∂n JK
i =1
F ∂U I
GH ∂n JK
F ∂G I
 GH ∂n JK
i
dni
T , p, n j
(11.34)
406
Engineering Thermodynamics
\ Equation (11.34) becomes
K
dG = Vdp – SdT +
 mi dni
i =1
Similar equations can be obtained for changes in H and F. Thus
K
dU = TdS – pdV +
 mi dni
i =1
K
dG = Vdp – SdT +
 mi dni
i =1
K
dH = TdS + Vdp +
 mi dni
(11.35)
i =1
K
dF = – SdT – pdV +
 mi dni
i =1
where
mi =
FG ∂U IJ
H ∂n K
i
=
S,V ,nj
FG ∂G IJ
H ∂n K
i
=
FG ∂H IJ
H ∂n K
i
T , p, n j
=
S , p, n j
FG ∂F IJ
H ∂n K
i
(11.36)
T ,V ,n j
Chemical potential is an intensive property.
Let us consider a homogeneous phase of a multi-component system, for which
K
dU = TdS – pdV +
 mi dni
i =1
If the phase is enlarged in size, U, S, and V will increase, whereas T, p and m will
remain the same. Thus
DU = TDS – pDV + Smi Dni
Let the system be enlarged to K-times the original size. Then
DU = KU – U = (K – 1)U
DS = KS – S = (K – 1)S
DV = (K – 1)V
Dni = (K – 1) ni
Substituting
(K – 1) U = T (K – 1) S – p (K – 1) V + Smi (K – 1) ni
\
U = TS – pV + S m ini
\
GT, p = S m ini
(11.37)
Let us now find a relationship if there is a simultaneous change in intensive property. Differentiating Equation (11.37)
Thermodynamic Relations, Equilibrium and Stability
407
dG = S ni dmi + S mi dni
(11.38)
at constant T and p, with only m changing.
When T and p change
dG = – SdT + Vdp + Smi dni
(11.39)
Combining Equations (11.38) and (11.39)
– SdT + Vdp – S ni d mi = 0
(11.40)
This is known as Gibbs-Duhem equation, which shows the necessary relationship
for simultaneous changes in T, p, and m.
Now
GT ◊ p = S mini = m1n1 + m2n2 ... + mKnK
For a phase consisting of only one constituent
G = mn
G
=g
n
i.e. the chemical potential is the molar Gibbs function and is a function of T and p
only.
For a single phase, multi-component system, m i is a function of T, p, and the
mole fraction xi.
m=
or
11.12
CONDITIONS OF EQUILIBRIUM OF A
HETEROGENEOUS SYSTEM
Let us consider a heterogeneous system of volume V, in which several homogeneous phases (f = a, b, ..., r) exist in equilibrium. Let us suppose that each phase
consists of i (= 1, 2, ..., C) constituents and that the number of constituents in any
phase is different from the others.
Within each phase, a change in internal energy is accompanied by a change in
entropy, volume and composition, according to
C
dUf = Tf dSf – pf dVf +  (mi dni)f
i =1
A change in the internal energy of the entire system can, therefore, be
expressed as follows:
r
r
r
r
C
f =a
f =a
f =a
f = a i =1
 dUf =  Tf dSf –  pf dVf +   (mi dni)f
(11.41)
Also, a change in the internal energy of the entire system involves changes in the
internal energy of the constituent phases.
r
dU = dUa + dU b + ... + dUr = Â dUf
f=a
Likewise, changes in the volume, entropy, or chemical composition of the
entire system result from contributions from each of the phases
r
dV = dVa + dVb + ... + dVr = Â dVf
f=a
408
Engineering Thermodynamics
r
dS = dSa + dSb + ... + dSr = Â dSf
f =a
r
dn = dna + dnb + ... + dnr = Â dnf
f =a
In a closed system in equilibrium, the internal energy, volume, entropy, and mass
are constant.
\
dU = dV = dS = dn = 0
dUa = – (dUb + ... + dUr) = -  dUj
or
j
dVa = - Â dVj
j
dSa = - Â dSj
(11.42)
j
dna = - Â dnj
j
where subscript j includes all phases except phase a.
Equation (11.41) can be written in terms of j independent variables and the dependent variable a (Equation 11.42).
(Ta dSa +  Tj dSj) – (pa dVa +  pj dVj
j
j
+ [ Â (mi dni)a + Â Â (m i dni) j ] = 0
i
j
j
Substituting from Equation (11.34)
(– Ta  dSj +  Tj dSj) – (– pa  dVj +  pj ◊ dVj)
j
j
j
j
+ [–   mia dnij +   (mi dni)j] = 0
j
t
j
i
where subscript i a refers to component i of phase a.
Rearranging and combining the coefficients of the independent variables, dSj,
dVj and dnj, gives
 (Tj – Ta)dSj –  ( pj – pa)dVj +   (mij – mia) dnij = 0
j
j
j
i
But since dSj, dVj, and dnj are independent, their coefficients must each be equal to
zero.
\
Tj = Ta, pj = pa, mij = mia
(11.43)
These equations represent conditions that exist when the system is in thermal, mechanical, and chemical equilibrium. The temperature and pressure of phase a must
be equal to those of all other phases, and the chemical potential of the ith component in phase a must be equal to the chemical potential of the same component in all
other phases.
Thermodynamic Relations, Equilibrium and Stability
11.13
409
GIBBS PHASE RULE
Let us consider a heterogeneous system of C chemical constituents which do not
combine chemically with one another. Let us suppose that there are f phases, and
every constituent is present in each phase. The constituents are denoted by subscripts and the phases by superscripts. The Gibbs function of the whole heterogeneous system is
C
C
C
i =1
i =1
i =1
GT, p = Â ni (1) mi(1) + Â ni(2) m i(2) + ... + Â ni(f) mi(f)
G is a function of T, p, and the n’s of which there are Cf in number. Since there are
no chemical reactions, the only way in which the n’s may change is by the transport
of the constituents from one phase to another. In this case the total number of moles
of each constituent will remain constant.
n1(1) + n1(2) + ... + n(f)
1 = constant
n2(1) + n2(2) + ... + n2(f) = constant
...................................................
nC(1) + nC(2) + ... + nC(f) = constant
These are the equations of constraint.
At chemical equilibrium, G will be rendered a minimum at constant T and p,
subject to these equations of constraint. At equilibrium, from Equation (11.43).
mij = mia
(2)
(f)
m (1)
1 = m1 = ... = m 1
m2(1) = m 2(2) = ... = m2(f)
(11.44)
...........................
m C(1) = m C(2) = ... = mC(f)
These are known as the equations of phase equilibrium. The equations of the phase
equilibrium of one constituent are (f – 1) in number. Therefore, for C constituents,
there are C (f – 1) such equations.
When equilibrium has been reached, there is no transport of matter from one
phase to another. Therefore, in each phase, S x = 1. For f phases, there are f such
equations available.
The state of the system at equilibrium is determined by the temperature, pressure, and Cf mole fractions. Therefore
Total number of variables = Cf + 2.
Among these variables, there are C(f – 1) equations of phase equilibrium and f
equations of Sx = 1 type. Therefore
Total number of equations = C (f – 1) + f
If the number of variables is equal to the number of equations, the system is
nonvariant. If the number of variables exceeds the number of equations by one,
then the system is called monovariant and is said to have a variance of 1.
410
Engineering Thermodynamics
The excess of variables over equations is called the variance, f. Thus
or
f = (Cf + 2) – [C(f – 1) + f]
f=C–f+2
(11.45)
This is known as the Gibbs Phase Rule for a non-reactive system. The variance ‘f ’
is also known as the degree of freedom.
For a pure substance existing in a single phase, C = 1, f = 1, and therefore, the
variance is 2. There are two properties required to be known to fix up the state of
the system at equilibrium.
If C = 1, f = 2, then f = 1, i.e. only one property is required to fix up the state of
a single-component two-phase system.
If C = 1, f = 3, then f = 0. The state is thus unique for a substance; and refers to
the triple point where all the three phases exist in equilibrium.
11.14
TYPES OF EQUILIBRIUM
The thermodynamic potential which
Reservoir
controls equilibrium in a system depends
T
on the particular constraints imposed on
dQ
the system. Let d-Q be the amount of heat
transfer involved between the system and
System
the reservoir in an infinitesimal irreversible process (Fig. 11.11). Let dS
Fig. 11.11 Heat Interaction between a
denote the entropy change of the system
System and its Surroundings
and dS0 the entropy change of the
reservoir. Then, from the entropy
principle
dS0 + dS > 0
d-Q
Since
dS0 = –
T
d-Q
–
+ dS > 0
T
d-Q – TdS < 0
or
During the infinitesimal process, the internal energy of the system changes by an
amount dU, and an amount of work pdV is performed. So, by the first law
d-Q = dU + pdV
Thus the inequality becomes
dU + pdV – TdS < 0
(11.46)
If the constraints are constant U and V, then the Equation (11.46) reduces to
dS > 0
The condition of constant U and V refers to an isolated system. Therefore, entropy
is the critical parameter to determine the state of thermodynamic equilibrium of an
isolated system. The entropy of an isolated system always increases and reaches a
maximum value when equilibrium is reached.
411
Thermodynamic Relations, Equilibrium and Stability
If the constraints imposed on the system are constant T and V, the Eq. (11.40)
reduces to
dU – d (TS) < 0
or
d(U – TS) < 0
\
dF < 0
which expresses that the Helmholtz function decreases, becoming a minimum at the
final equilibrium state.
If the constraints are constant T and p, the Eq. (11.46) becomes
dU + d(pV) – d(TS) < 0
d(U + pV – TS) < 0
\
dG < 0
The Gibbs function of a system at constant T and p decreases during an irreversible
process, becoming a minimum at the final equilibrium state. For a system constrained in a process to constant T and p, G is the critical parameter to determine the
state of equilibrium.
The thermodynamic potential and the corresponding constrained variables are
shown below.
S
U
V
H
F
P
G
T
This trend of G, F, or S establishes four types of equilibrium, namely (a) stable, (b)
neutral, (c) unstable, and (d) metastable.
A system is said to be in a state of stable equilibrium if, when the state is perturbed, the system returns to its original state. A system is not in equilibrium if there
is a spontaneous change in the state. If there is a spontaneous change in the system,
the entropy of the system increases and reaches a maximum when the equilibrium
condition is reached (Fig. 11.12). Both A and B (Fig. 11.13) are assumed to be at
the same temperature T. Let there be some spontaneous change; the temperature of
A rises to T + dT1, and that of B decreases to T – dT2. For simplicity, let the heat
capacities of the bodies be the same, so that dT1 = dT2 = dT. If d-Q is the heat
interaction involved, then the entropy change
d-Q
d- Q
dSA =
, dSB =
T + dT
T - dT
\
dS = dSA + dSB = d- Q
LM 1 − 1 OP = − 2 ⋅ d T ◊ d Q
NT + dT T − dT Q T
-
2
So there is a decrease in entropy for the isolated system of A and B together. It is
thus clear that the variation in temperature dT cannot take place. The system, therefore, exists in a stable equilibrium condition. Perturbation of the state produces an
absurd situation and the system must revert to the original stable state. It may be
observed:
If for all the possible variations in state of the isolated system, there is a negative
change in entropy, then the system is in stable equilibrium.
412
Engineering Thermodynamics
Diathermal wall
le
le
ib
S
s
dQ B
Im
po
ss
Po
sib
T
A
T
Equilibrium
Isolated system
t
Fig. 11.12
Possible Process for
an Isolated System
(dS)U,V > 0
(dS )U,V = 0
(dS )U,V < 0
Fig. 11.13
Spontaneous Changes in
A and B due to Heat
Interaction
Spontaneous change
Equilibrium
Criterion of stability
(11.47)
Similarly
(dG)p,T < 0, (dF) T,V < 0
Spontaneous change
(dG)p,T = 0, (dF) T,V = 0
Equilibrium
(11.48)
(dG)p,T > 0, (dF ) T ◊ V > 0
Criterion of stability
A system is in a state of stable equilibrium if, for any finite variation of the
system at constant T and p, G increases, i.e. the stable equilibrium state corresponds to the minimum value of G.
A system is said to be in a state of neutral equilibrium when the thermodynamic
criterion of equilibrium (G, F, S, U, or H ) remains at constant value for all possible
variations of finite magnitude. If perturbed, the system does not revert to the original state.
For a system at constant T and p, the criterion of neutral equilibrium is
d GT,p = 0
Similarly
d FT,V = 0, d HS, p = 0, d US,V = 0, d SU,V = 0
A system is in a state of unstable equilibrium when the thermodynamic
criterion is neither an extremum nor a constant value for all possible variations in
the system. If the system is in an unstable equilibrium, there will be a spontaneous
change accompanied by
dGT,p < 0, dF T,V < 0, dUS,V < 0, d HS,p < 0, dSU,V > 0
A system is in a state of metastable equilibrium if it is stable to small but not to
large disturbances. A mixture of oxygen and hydrogen is in a metastable equilibrium. A little spark may start a chemical reaction. Such a mixture is not in its most
stable state, even though in the absence of a spark it appears to be stable.
Figure 11.14 shows different types of equilibrium together with their mechanical
analogies. S has been used as the criterion for equilibrium.
413
S
S
S
S
Thermodynamic Relations, Equilibrium and Stability
Unstable
Stable
Neutral
Composition
Metastable
Fig. 11.14 Types of Equilibrium
11.15
LOCAL EQUILIBRIUM CONDITIONS
Let an arbitrary division of an isolated system be considered, such that
S = S1 + S2, U = U1 + U2
Then for equilibrium, it must satisfy the condition
(dS)U,V = 0
to first order in small displacements (otherwise d S could be made positive because
of higher order terms). Now to the first order in a very small change
d 1S =
FG ∂S IJ dU + FG ∂S IJ dU + FG ∂S IJ dV + FG ∂S IJ dV
H ∂U K
H ∂U K
H ∂V K
H ∂V K
1
2
1 V
2
Now
TdS = dU + pdV
\
FG ∂S IJ = 1 , FG ∂S IJ = p
H ∂U K T H ∂ V K T
V
\
1
1 U
V
2
U
U
d 1S =
1
1
p
p
dU1 +
dU2 + 1 dV1 + 2 dV2
T1
T1
T2
T2
Again
dU1 = – dU2 and dV1 = – dV2
\
d1 S =
FG 1 − 1 IJ dU + FG p − p IJ dV + Second order terms
HT T K
HT T K
1
2
1
2
1
1
When
2
2
d1 S = 0, at equilibrium
T1 = T2, p1 = p2
1
414
11.16
Engineering Thermodynamics
CONDITIONS OF STABILITY
At equilibrium, S = Smax , F = Fmin, G = Gmin, and dS = 0, dF = 0; dG = 0; these are
necessary but not sufficient conditions for equilibrium. To prove that S is a maximum, and G or F a minimum, it must satisfy
d2S < 0, d2F > 0, d 2G > 0
If the system is perturbed, and for any infinitesimal change of the system
(dS)U,V < 0, (dG)p,T > 0, (dF )T,V > 0
it represents the stability of the system. The system must revert to the original state.
For a spontaneous change, from Eq. (11.46)
dU + pdV – Td S < 0
For stability
dU + pd V – Td S > 0
Let us choose U = U (S, V ) and expand d U in powers of d V and d S.
FG ∂U IJ dS + 1 FG ∂ U IJ (dS) + FG ∂U IJ dV
H ∂S K
2 H ∂S K
H ∂V K
1 F∂ UI
∂ U
+ G
(dV ) +
◊ dV ◊ dS +
J
2 H ∂V K
∂V ⋅ ∂ S
1 Fδ UI
= Td S – pdV + G
J (dS)
2 H δS K
1F∂ UI
∂U
+ G
(dV) +
dV ◊ dS +
J
2 H ∂V K
∂V ◊ ∂ S
2
dU =
2
2
V
S
V
2
2
2
2
S
2
2
2
2
V
2
2
2
S
The third order and higher terms are neglected.
Since dU + pdv – TdS > 0, it must satisfy the conditions given below
FG ∂ U IJ > 0, FG ∂ U IJ > 0, ∂ U > 0
H ∂ S K H ∂V K ∂V ◊ ∂ S
2
2
2
2
2
V
S
These inequalities indicate how the signs of some important physical quantities
become restricted for a system to be stable.
Since
FG ∂U IJ = T
H ∂S K
FG ∂ U IJ = FG ∂T IJ = T
H ∂S K H ∂S K C
V
2
2
V
V
\
Since T > 0 K
\
v
T
>0
Cv
Cv > 0
(11.49)
Thermodynamic Relations, Equilibrium and Stability
415
which is the condition of thermal stability.
Also
FG ∂U IJ = – p
H ∂V K
FG ∂ U IJ = −FG ∂ p IJ
H ∂ V K H ∂V K
FG ∂ p IJ < 0
H ∂V K
S
2
2
S
\
S
(11.50)
S
i.e. the adiabatic bulk modulus must be negative.
Similarly, if F = F (T, V), then by Taylor’s expansion, and using appropriate
substitution
d F = – SdT – pdV +
+
FG
H
IJ
K
1 ∂2 F
(dV)2
2
2 ∂V T
FG IJ
H K
1 ∂2 F
∂ 2 ⋅F
(dT )2 +
◊dV ◊ d T +
2
2 ∂T S
∂V ⋅ ∂T
For stability
dF + Sd T + pdV > 0
FG ∂ F IJ > 0
H ∂V K
2
2
T
Again
FG ∂ F IJ = – p
H ∂V K
FG ∂ F IJ = – FG ∂ p IJ
H ∂V K H ∂V K
FG ∂ p IJ < 0
H ∂V K
T
2
\
2
T
\
T
(11.51)
T
which is known as the condition of mechanical stability. The isothermal bulk modulus must also be negative.
Solved Examples
Example 11.1
(a) Derive the equation
FG ∂C IJ = – T Ê ∂ V ˆ
Á
˜
H ∂p K
Ë ∂T ¯
2
p
2
T
p
416
Engineering Thermodynamics
(b) Prove that Cp of an ideal gas is a function of T only.
(c) In the case of a gas obeying the equation of state
pv
= 1 + B ¢p
RT
where B¢ is a function of T only, show that
C p = R Tp
d2
dT 2
(B¢T) + (Cp)0
where (Cp)0 is the value at very low pressures.
Solution
(a)
FG ∂ S IJ
H ∂T K
FG ∂C IJ = T FG ∂ S IJ
H ∂ p ⋅ K H ∂T ◊ ∂p K
FG ∂S IJ = – FG ∂V IJ , by Maxwell’s relation
H ∂ p K H ∂T K
F ∂ V IJ
∂ S
=–G
∂ p ⋅ ∂T
H ∂T K
FG ∂C IJ = – T FG ∂ V IJ
H ∂pK
H∂T K
Cp = T
p
2
p
T
Now
p
T
2
\
2
2
p
\
2
p
2
T
p
(b) For an ideal gas
V=
nRT
p
FG ∂V IJ = nR and FG ∂ V IJ = 0
H ∂T K p H ∂T K
F ∂C I = 0, i.e. C is a function of T alone.
GH ∂ p JK
2
2
p
\
p
p
p
T
(c)
pv
= 1 + B¢p
RT
\
B¢p =
pv
–1
RT
B¢T =
T pv
v T
−1 =
−
p RT
R p
FG
H
IJ FG
K H
IJ
K
Proved.
Thermodynamic Relations, Equilibrium and Stability
LM ∂ ( B′ T )OP = + 1 FG ∂v IJ − 1
N∂ T Q R H ∂ T K p
LM ∂ a B′ T f OP = 1 F ∂ v I = − 1 F ∂C I
N ∂ T Q R GH ∂T JK RT GH ∂ p JK
p
417
p
2
2
p
2
2
p
p
T
\ On integration
Cp = – RTp
d2
(B¢T ) + C p0
dT 2
where C p0 (integration constant) is the value of Cp at very low values of pressure.
Example 11.2 The Joule-Kelvin coefficient mJ is a measure of the temperature
change during a throttling process. A similar measure of the temperature change
produced by an isentropic change of pressure is provided by the coefficient ms,
where
∂T
mS =
∂p S
Prove that
V
m S – mJ =
Cp
FG IJ
H K
Solution
The Joule-Kelvin coefficient, mj, is given by
T
FG ∂V IJ − V
H ∂T K
p
Cp
Since Cp = T
FG ∂S IJ and by Maxwell’s relation
H ∂TK
FG ∂V IJ = - FG ∂S IJ
H ∂ T K H ∂ pK
F ∂S IJ
-T G
H ∂ pK - V
m =
F ∂ S IJ C
TG
H ∂T K
F ∂S IJ FG ∂T IJ - V
m =–G
H ∂ pK H ∂ SK C
FG ∂S IJ FG ∂T IJ FG ∂ p IJ = – 1
H ∂ pK H ∂ SK H ∂ T K
p
p
T
T
J
p
p
\
J
T
Since
T
p
p
p
s
418
Engineering Thermodynamics
\
mJ = + m S –
\
ms – mJ =
V
Cp
V
Cp
Proved.
Alternative method:
From the second Tds equation
FG ∂V IJ dp
H ∂T K
FG ∂T IJ = m = T FG ∂ V IJ
H ∂pK
C H ∂T K
O
1 L F ∂V I
M
TG J − VP
m =
C M H ∂T K
PQ
N
TdS = Cp dT – T
p
s
p
s
Now
p
J
p
\
ms – mJ =
p
V
Cp
Proved.
Example 11.3 If the boiling point of benzene at 1 atm pressure is 353 K, estimate the approximate value of the vapour pressure of benzene at 303 K.
Solution
Using Clapeyron’s equation and Trouton’s rule, Eq. (11.21),
RS 88 F1 − T I UV
T R H T KW
R 88 F1 − 353I UV = 17.7 kPa
= 101.325 exp S
T 8.3143 H 303K W
p = 101.325 exp
Example 11.4
given by
B
The vapour pressure, in mm of mercury, of solid ammonia is
ln p = 23.03 –
3754
T
and that of liquid ammonia by
3063
T
(a) What is the temperature of the triple point? What is the pressure? (b) What
are the latent heats of sublimation and vaporization? (c) What is the latent heat of
fusion at the triple point?
ln p = 19.49 –
Solution
At the triple point, the saturated solid and saturated liquid lines meet.
23.03 –
\
3754
3063
= 19.49 –
T
T
T = 195.2 K
Thermodynamic Relations, Equilibrium and Stability
419
ln p = 23.03 – 3754
195.2
ln p = 3.80
p = 44.67 mm Hg
With the assumptions, v¢¢¢ >> v¢ and v¢¢¢ ~
RT
,
p
Clausius-Clapeyron equation reduces to
dp
p
=
◊ lsub
RT 2
dT
where lsub is the latent heat of sublimation.
The vapour pressure of solid ammonia is given by
3754
ln p = 23.03 –
T
3754
1 dp
\
⋅
=
T2
p dT
\
\
dp
p
p
= 3754 2 =
◊lsub
T
RT2
dT
lsub = 3754 ¥ 8.3143 = 31,200 kJ/kg mol
The vapour pressure of liquid ammonia is given by
ln p = 19.49 –
3063
T
dp
p
p
= 3063 2 =
lvap
dT
T
RT 2
where lvap is the latent heat of vaporization.
\
lvap = 3063 ¥ 8.3143 = 25,500 kJ/kg mol
At the triple point
lsub = lvap + lfu
where lfu is the latent heat of fusion.
\
\
lfu = lsub – lvap
= 31,200 – 25,500 = 5,700 kJ/kg mol
Example 11.5 Explain why the specific heat of a saturated vapour may be negative.
Solution As seen in Fig. 11.15, if heat is transferred along the saturation line,
there is a decrease in temperature. The slope of the saturated vapour line is negative, i.e. when dS is positive, dT is negative. Therefore, the specific heat at constant
saturation
C¢¢¢
sat = T
FG d S′′′ IJ
H dT K
420
Engineering Thermodynamics
T
Saturated vapour line
c¢¢¢
T
dQ
s
ds
Fig. 11.15
is negative. From the second TdS equation
∂V
TdS = CpdT – T
∂T
FG IJ dp
H K
F ∂V ¢¢¢ IJ FG d p IJ
d S¢¢¢
T
=C –TG
dT
H ∂T K H dT K
p
p
p
sat
lvap
nR
= Cp – T
⋅
p T V ′′′ − V ′′
a
f [using pV¢¢¢ = nRT and
Clapeyron’s equation]
C¢¢¢
sat = Cp –
\
C¢¢¢
sat = Cp –
lvap
V ′′′
⋅
T V ′′′
lvap
[∵ V¢¢¢ >> V¢¢]
T
Now the value of lvap /T for common substances is about 83.74 J/kg mol K
(Trouton’s rule), where Cp is less than 41.87 J/g mol K. Therefore, C¢¢¢
sat can be
negative.
Proved.
Example 11.6 (a) Establish the condition of equilibrium of a closed composite
system consisting of two simple systems separated by a movable diathermal wall
that is impervious to the flow of matter.
(b) If the wall were rigid and diathermal, permeable to one type of material, and
impermeable to all others, state the condition of equilibrium of the composite system.
(c) Two particular systems have the following equations of state
N
1
3 N p
= R 1, 1= 1 R
T1
2 U 1 T1 V1
and
N
1
3 N p
= R 2 , 2 =R 2
T2
2 U 2 T2
V2
where R = 8.3143 kJ/kg mol K, and the subscripts indicate systems 1 and 2. The
mole number of the first system is N1 = 0.5, and that of the second is N2 = 0.75. The
two systems are contained in a closed adiabatic cylinder, separated by a movable
diathermal piston. The initial temperatures are T1 = 200 K and T2 = 300 K, and the
421
Thermodynamic Relations, Equilibrium and Stability
total volume is 0.02 m3. What is the energy and volume of each system in equilibrium? What is the pressure and temperature?
Solution
For the composite system, as shown in Fig. 11.16
U1 + U2 = constant
V1 = V2 = constant
Movable diathermal
wall impervious to
the flow of matter
Subsystem-1
Subsystem-2
(a)
Subsystem-1
Subsystem-2
1
Rigid and diathermal
wall permeable to flow of
one type of material
2
Movable, diathermal
wall
(c)
(b)
Fig. 11.16
The values of U1, U2, V1, and V2 would change in such a way as to maximize the
value of entropy. Therefore, when the equilibrium condition is achieved
dS = 0
for the whole system. Since
S = S1 + S2
= S1(U1, V1, ..., Nk1 ...) + S2 (U2, V2, ..., Nk2 ...)
\
dS =
FG ∂S IJ
H ∂U K
1
1
+
FG ∂S IJ
H ∂U K
2
2
dU1 +
V2 …, N k 2 …,
=
1
dV1
1 U …, N …,
k1
1
V1 …, N k 1 …
dU2 +
FG ∂S IJ
H ∂V K
FG ∂S IJ
H ∂V K
2
2
dV2
V2 …, Nk 2 …
p
p
1
1
dU1 + 1 dV1 +
dU2 + 2 dV2
T1
T2
T1
T2
Since dU1 + dU2 = 0 and dV1 + dV2 = 0
422
Engineering Thermodynamics
\
dS =
FG 1 − 1 IJ dU + FG p − p IJ dV = 0
HT T K
HT T K
1
2
1
2
1
1
1
2
Since the expression must vanish for arbitrary and independent values of dU1
and dV1
p
p
1
1
−
= 0 and 1 − 2 = 0
T1 T2
T1 T2
or
p1 = p2 and T1 = T2
\
These are the conditions of mechanical and thermal equilibrium.
(b) We will consider the equilibrium state of two simple subsystems (Fig. 11.16
(b)) connected by a rigid and diathermal wall, permeable to one type of material
(N1) and impermeable to all others (N2, N3, ... Nr). We thus seek the equilibrium
values of U1 and U2, and of N1–1 and N 1–2 (i.e. material N1 in subsystems 1 and 2
respectively.)
At equilibrium, an infinitesimal change in entropy is zero
dS = 0
Now
dS = dS1 + dS2
=
FG ∂S IJ
H ∂U K
1
dU 1 +
1 V ,N −
1 1 1,…
+
FG ∂S IJ
H ∂U K
2
2
FG ∂S IJ
H ∂N K
FG ∂S IJ
H ∂N K
1
1-1 U , V , N 1 1
1 2 ,…
2
dU2 +
1- 2
V2 , N1-2
dN1 – 1
dN1 – 2
U2 , V2 , N 2 -2
From the equation
TdS = dU + pdV – mdN
FG ∂ S IJ
H ∂U K
\
and
=
V, N ,…
FG IJ
H K
1 ∂S
µ
,
=–
T ∂ N U ,V
T
dN1 – 1 + dN1 – 2 = 0
dU1 + dU2 = 0
dS =
FG 1 − 1 IJ dU – FG µ − µ IJ dN
HT T K
H T T K
1−1
1− 2
1
2
1
1
2
As dS must vanish for arbitrary values of both dU1 and dN1 – 1
T1 = T2
m1 – 1 = m1 – 2
which are the conditions of thermal and chemical equilibrium.
1– 1 = 0
Thermodynamic Relations, Equilibrium and Stability
(c)
423
N1 = 0.5 g mol,
N2 = 0.75 g mol
T1 – 1 = 200 K, T1 – 2 = 300 K
V = V1 + V2 = 0.02 m 3
U1 + U2 = constant
DU1 + DU2 = 0
Let Tf be the final temperature (Fig. 11.16(c))
(Uf – 1 – Ui – 1) = – (Uf – 2 – Ui – 2)
3
3
RN1 (Tf – Ti – 1) = – RN 2 (Tf – Ti – 2)
2
2
0.5 (Tf – 200) = – 0.75 (Tf – 300)
\
1.25 Tf = 325
or
Tf = 260 K
Uf – 1 =
3
3
R N1Tf = ¥ 8.3143 ¥ 0.5 ¥ 10–3 ¥ 260 = 1.629 kJ
2
2
Uf – 2 =
3
¥ 8.3143 ¥ 0.75 ¥ 10–3 ¥ 260 = 2430 kJ
2
Vf-2 =
RN1 T f -1
p f -1
Vf – 2 =
Vf – 1 + Vf – 2 =
At equilibrium
p f - 1 = p f -2 = p
f
RN 2 T f − 2
p f −2
RTf
pf
Tf – 1 = Tf – 2 = Tf
(N1 + N2 ) = 0.02 m 3
8.3143 × 260 ¥ 1.25 ¥ 10–3 = 0.02 m3
pf
\
pf =
8.3143 ¥ 260 ¥ 1.25 ¥ 10 -3
kN/m2
0.02
= 135 kN/m2 = 1.35 bar
8.3143 × 0.5 × 10 −3 × 260
= 0.008 m3
135
\
Vf – 1 =
\
Vf – 2 = 0.02 – 0.008 = 0.012 m3
Example 11.7
(a)
Show that for a van der Waals’ gas
FG ∂c IJ = 0
H ∂v K
v
T
424
Engineering Thermodynamics
v2 - b
v1 - b
(b) (s2 – s1)T = R ln
(c) T (v – b)R/cv = constant, for an isentropic
(d) cp – cv =
R
1 - 2a ( v - b ) / RT v3
2
Ê 1
1ˆ
(e) (h2 – h1)T = (p2 v2 – p1 v1) + a Á
- ˜
Ë v1 v2 ¯
Solution
(a) From the energy Eq. (11.13)
FG ∂U IJ = T FG ∂ p IJ – p
H ∂V K H ∂T K
F ∂ p IJ + FG ∂ p IJ − FG ∂ p IJ
∂U
=TG
∂V ⋅ ∂T
H ∂T K H ∂ T K H ∂ T K
∂ U
F ∂ p IJ
=TG
∂V ⋅ ∂ T
H ∂T K
F ∂U IJ
C =G
H ∂T K
FG ∂C IJ = ∂ U = T F ∂ p I = F ∂c I
H ∂V K ∂T ◊ ∂V GH ∂T JK GH ∂v JK
T
V
2
2
2
V
V
2
V
2
2
V
v
V
2
v
2
v
2
T
V
T
For a van der Waals’ gas
F p + a I (v – b) = RT
H vK
2
p=
RT
a
−
v − b v2
FG ∂ p IJ = 0
H ∂T K
FG ∂c IJ = 0
H ∂v K
2
∵
2
V
\
v
Proved (a)
T
\
cv is independent of volume.
(b) From the first Tds Eq. (11.8)
Tds = cv dT + T
FG ∂p IJ dv
H ∂T K
V
Thermodynamic Relations, Equilibrium and Stability
and energy Eq. (11.13),
425
FG ∂U IJ = T FG ∂p IJ –p
H ∂V K H ∂T K
dT 1 L F ∂U I O
ds = c
+ M p +G
J P dv
T
T N H ∂V K Q
T
V
v
T
For van der Waals’ gas
FG ∂U IJ = a
H ∂V K v
2
T
\
ds = cv
= cv
\
F
H
I
K
dT 1
a
+
p + 2 dv
T T
v
dT
R
+
dv
T v-b
(s2 – s1)T = R ln
v2 - b
v1 - b
Proved (b)
(c) At constant entropy
cv
or
dT
R
+
dv = 0
T v-b
dT R d v
+
=0
T cv v - b
by integration,
T (v – b)R/Cv = constant
(d)
cp – cv = T
Proved (c)
FG ∂p IJ FG ∂ v IJ
H ∂T K H ∂ T K
LF ∂U IJ + pOP FG ∂v IJ
= MG
NH ∂V K Q H ∂T K
a
I F ∂v I
=F
H v + pK GH ∂T JK
F RT IJ FG ∂v IJ
=G
H v - b K H ∂T K
V
p
T
p
2
p
p
From the equation
F p + a I (v – b) = RT
H vK
2
426
Engineering Thermodynamics
\
(v – b) (– 2av–3)
FG ∂v IJ + F p + a I FG ∂v IJ = R
H ∂T K H v K H ∂T K
2
p
p
FG ∂v IJ = R /( v - b )
H ∂T K RT - 2a
p
\
cp – cv =
(v - b)
v3
R
Proved (d)
a f
FG ∂U IJ = T FG ∂p IJ – p = a
H ∂V K H ∂T K
v
(e)
2
1 - 2a v - b / RTv 3
2
T
duT =
\
V
a
dvT
v2
(u2 – u1)T = a
FG 1 - 1 IJ
Hv v K
1
\
2
(h2 – h1)T = (p2v2 – p1v1) + a
FG 1 - 1 IJ
Hv v K
1
Example 11.8
Proved (e)
2
The virial equation of state of a gas is given by
pv = RT (1 + B¢p + C¢p2 + … )
Show that
È Ê ∂v ˆ
˘
d B¢
lim ÍT Á ˜ - v ˙ = RT2
p Æ0 Í Ë ∂ T ¯
dT
˙˚
p
Î
Hence, prove that the inversion temperature of a van der Waals’ gas is twice the
Boyle temperature.
Solution
pv = RT (1 + B¢p + C¢p2 + … )
v=
RT
+ RTB¢ + RTpC¢ + …
p
FG ∂v IJ = R + RT d B¢ + RB¢ + RTp dC ¢ + RpC¢ + …
H ∂T K p
dT
dT
p
\
T
FG ∂v IJ = RT + RT d B¢ + RTB¢ + RT p dC ¢ + RTpC¢ + …
H ∂T K p
dT
dT
d B¢
dC¢
F ∂v I
+…
T G J – v = RT
+ RT p
H ∂T K
dT
dT
O
1 L F ∂v I
M
T G J - vP
m =
c M H ∂T K
PQ
N
2
2
p
\
2
J
p
2
p
Thermodynamic Relations, Equilibrium and Stability
\
=
LM
N
OP
Q
RT 2 d B¢
dC ¢
+p
+…
c p dT
dT
lim mJ =
pÆ0
RT 2 d B ¢
c P dT
LM F d v I - vOP = RT d B¢
dT
MN GH dT JK PQ
2
lim T
or
427
pÆ 0
Proved
p
For a van der Waals’ gas, to find Boyle temperature TB,
a
B=b–
=0
RT
a
\
TB =
bR
B
b
a
=
RT
RT R 2 T 2
dB¢
2a
b
=+ 2 3
2
RT
R T
dT
B¢ =
\
lim mJ =
pÆ0
F
H
RT 2
b
2a
+
cp
RT 2 R2 T 3
I =0
K
2a
b
= 2 3
RT 2
R T
2a
\
Ti =
bR
\
T1 = 2TB
or Inversion temperature = 2 ¥ Boyle temperature
Proved
Example 11.9 Over a certain range of pressures and temperatures the equation
of a certain substance is given by the relation
\
v=
RT C
p T3
where C is a constant. Derive an expression for: (a) the change of enthalpy and (b)
the change of entropy, of this substance in an isothermal process.
Solution
(a) From Eq. (11.15)
LM
MN
FG ∂v IJ OP dp
H ∂T K PQ
LMv - T FG ∂v IJ OPdp
MN H ∂T K PQ
dh = cp dT + v - T
\
(h2 – h1)T =
Now,
v=
z
2
1
RT C
- 3
p
T
p
p
428
Engineering Thermodynamics
FG ∂v IJ = R + 3C
H ∂T K p T
F ∂v I RT + 3C
TG J =
H ∂T K p T
F ∂v I RT - C - RT - 3C = 4C
v–T G J =
H ∂T K p T p T T
4
p
\
3
p
\
3
3
3
p
on substitution,
(h2 – h1)T =
z
p2
p1
-
4C
dp
T3
4C
(p1 – p2)T
T3
(b) Using second Tds equation
=
Tds = cp dT – T
FG ∂v IJ dp
H ∂T K
p
\
FG ∂v IJ dp
H ∂T K
F R 3C IJ dp
=G +
Hp T K
dsT = –
T
p
4
(s2 – s1)T = R ln
T
p1 3C
+
(p1 – p2)T
p2 T 4
Review Questions
11.1 What is the condition for exact differential?
11.2 Derive Maxwell’s equations.
11.3 Write down the first and second TdS equations, and derive the expression for
the difference in heat capacities, Cp and Cv. What does the expression signify?
11.4 Define volume expansivity and isothermal compressibility.
11.5 Show that the slope of an isentrope is greater than that of an isotherm on p–v
plot. How is it meaningful for estimating the work of compression?
11.6 What is the energy equation? How does this equation lead to the derivation of
the Stefan-Boltzman law of thermal radiation?
11.7 Show that the internal energy and enthalpy of an ideal gas are functions of
temperature only.
Thermodynamic Relations, Equilibrium and Stability
429
11.8 Why are dU = Cv dT and dH = Cp dT true for an ideal gas in any process,
whereas these are true for any other substance only at constant volume and at
constant pressure respectively?
11.9 Explain Joule-Kelvin effect. What is inversion temperature?
11.10 What is Joule-Thomson coefficient? Why is it zero for an ideal gas?
11.11 Why does the hydrogen gas need to be precooled before being throttled to get
the cooling effect?
11.12 Why does the maximum temperature drop occur if the state before throttling
lies on the inversion curve?
11.13 Why does the Gibbs function remain constant during phase transition?
11.14 What are the characteristics of the first order phase transition?
11.15 Write down the representative equation for phase transition. Why does the
fusion line for water have negative slope on the p–T diagram?
11.16 Why is the slope of the sublimation curve at the triple point on the p–T diagram greater than that of the vaporization curve at the same point?
11.17 Explain how thermodynamic properties are evaluated from an equation of
state.
11.18 Explain how enthalpy change and entropy change of a gas are estimated from
an equation of state.
11.19 State the important thermodynamic criteria which an equation of state must
satisfy.
11.20 Explain how Boyle temperature is yielded from
lim
pÆ0
FG ∂z IJ = 0
H ∂p K
T
11.21 What is foldback temperature?
11.22 Show that for an inversion curve
FG ∂z IJ = 0.
H ∂T K
p
11.23
11.24
11.25
11.26
11.27
11.28
11.29
11.30
11.31
11.32
11.33
Define chemical potential of a component in terms of U, H, and G.
What is the use of the Gibbs entropy equation?
Explain the significance of the Gibbs-Duhem equation.
State the conditions of equilibrium of a heterogeneous system.
What do you understand by phase equilibrium?
Give the Gibbs phase rule for a nonreactive system. Why is the triple point of
a substance nonvariant?
What are the four types of equilibrium? What is stable equilibrium?
State the conditions of spontaneous change, equilibrium and criterion of
stability for: (a) a system having constant U and V (i.e., isolated), and (b) a
system having constant T and p.
What do you understand by neutral and unstable equilibrium?
What is metastable equilibrium?
Show that for a system to be stable, these conditions are satisfied
(a) CV > 0 (thermal stability)
(b)
FG ∂p IJ < 0
H ∂V K
T
(mechanical stability)
430
Engineering Thermodynamics
Problems
11.1 Derive the following equations
FG ∂F IJ = – T FG ∂F / T IJ
H ∂T K V
H ∂T K V
F ∂2 F I
(b) C = – T G
H ∂T 2 JK v
F ∂G IJ = – T FG ∂G / T IJ
(c) H = G – T G
H ∂T K
H ∂T K
F ∂ GI
(d) C = – T G
H ∂T JK
2
(a) U = F – T
v
2
p
p
2
p
2
p
11.2 (a) Derive the equation
FG ∂c IJ = T FG ∂ p IJ
H ∂v K H ∂T K
2
v
2
T
V
(b) Prove that cv of an ideal gas is a function of T only.
(c) In the case of a gas obeying the equation of state
pv
B¢¢
=1+
RT
v
where B¢¢ is a function of T only, show that
RT d 2
(B¢¢ T) + (c v) 0
v dT 2
where (c v)0 is the value at very large volumes.
11.3 Derive the third TdS equation
cv = -
TdS = Cv
FG ∂T IJ dp + C FG ∂T IJ dV
H ∂V K
H ∂p K
p
v
p
and show that the three TdS equations may be written as
bT
dV
k
(b) TdS = Cp dT – Vb Tdp
(a) TdS = Cv dT +
(c) TdS =
Cp
Cv
kdp +
dV
b
bV
11.4 Derive the equations
∂V
∂p
(a) Cp = T
∂T p ∂T
Cp
∂p
(b)
=
∂T s VbT
FG IJ FG IJ
H K H K
FG IJ
H K
s
Thermodynamic Relations, Equilibrium and Stability
(c)
431
b∂p / ∂T g = g
b∂p / ∂T g g - 1
s
v
11.5 Derive the equations
∂p
∂V
(a) Cv = – T
∂T v dT
C k
∂V
(b)
=- v
∂T s
bT
FG IJ FG IJ
H KH K
FG IJ
H K
b∂V / ∂T g 1
(c) ∂V / ∂T = 1 - g
b
g
s
s
p
11.6 (a) Prove that the slope of a curve on a Mollier diagram representing a reversible isothermal process is equal to
1
T–
b
(b) Prove that the slope of a curve on a Mollier diagram representing a reversible isochoric process is equal to
g -1
T+
b
11.7 (a) Show that
∂V / T
mJ cp = T 2
∂T p
FG
H
IJ
K
For 1 mole of a gas, in the region of moderate pressures, the equation of sate
may be written as
pv
= 1 + B¢p + C¢p2
RT
where B¢ and C¢ are functions of temperature only.
(b) Show that as p Æ 0
dB¢
mJ cp Æ RT 2
dT
(c) Show that the equation of the inversion curve is
dB¢ / dT
dC ¢ / dT
11.8 Prove the following functional relationship of the reduced properties for the
inversion curve of a van der Waals gas
p=-
Tr =
b
g and p = 9b2v - 1g
3 3vr - 1
4 v r2
2
r
r
vr2
Hence, show that
Maximum inversion temperature
= 6.75
Critical temperature
Minimum inversion temperature
and
= 0.75
Critical temperature
432
Engineering Thermodynamics
11.9 Estimate the maximum inversion temperature of hydrogen if it is assumed to
obey the equation of state
pV = RT + B1p + B2 p2 + B3p3 +
For hydrogen, B1 ¥ 105 = a + 10–2 bT + 102 c/T
where
a = 166, b = – 7.66, c = – 172.33
11.10 The vapour pressure of mercury at 399 K and 401 K is found to be 0.988 mm
and 1.084 mm of mercury respectively. Calculate the latent heat of vaporization of liquid mercury at 400 K.
Ans. 61,634.96 kJ/kg mol
11.11 In the vicinity of the triple point, the vapour pressure of liquid ammonia (in
atmospheres) is represented by
3063
T
This is the equation of the liquid-vapour boundary curve in a p–T diagram.
Similarly, the vapour pressure of solid ammonia is
ln p = 15.16 –
3754
T
(a) What is the temperature and pressure at the triple point?
(b) What are the latent heats of sublimation and vaporization?
(c) What is the latent heat of fusion at the triple point?
Ans. 195.2 K, 0.585 atm., 1498 kJ/kg, 1836 kJ/kg, 338 kJ/kg
11.12 It is found that a certain liquid boils at a temperature of 95ºC at the top of a hill,
whereas it boils at a temperature of 105ºC at the bottom. The latent heat is
4.187 kJ/g mole. What is the approximate height of the hill?
Ans. 394 m
11.13 Show that for an ideal gas in a mixture of ideal gases
ln p = 18.70 –
m k - hk
dT + vk dp + RTd ln xk
T
11.14 Compute mj for a gas whose equation of state is
dm k =
p(v – b) = RT
Ans. mJ = – b/cp
11.15 Show that
FG ∂b IJ = - FG ∂k IJ
H ∂p K H ∂T K
F ∂u I
F ∂v I F ∂v I
(b) G J = - T G J - p G J
H ∂T K H ∂p K
H ∂p K
(a)
T
T
p
p
T
11.16 Two particular systems have the following equations of state
3 N (1)
1
5 N ( 2)
1
=
R
and
=
R
2 U (1)
T ( 2) 2 U (2)
T (1)
Thermodynamic Relations, Equilibrium and Stability
433
where R = 8.3143 kJ/kg mol K. The mole number of the first system is N (1) =
2, and that of the second is N (2) = 3. The two systems are separated by a
diathermal wall, and the total energy in the composite system is 25.120 kJ.
What is the internal energy of each system in equilibrium?
Ans. 7.2 kJ, 17.92 kJ
11.17 Two systems with the equations of state given in Problem 11.16 are separated
by a diathermal wall. The respective mole numbers are N(1) = 2 and N(2) = 3.
The initial temperatures are T(1) = 250 K and T(2) = 350 K. What are the values
of U(1) and U(2) after equilibrium has been established? What is the equilibrium temperature?
Ans. 8.02 kJ, 20.04 kJ, 321.4 K
11.18 Show that the change in latent heat L with temperature is given by the following relation
L v¢¢¢ b ¢¢¢ - v¢¢b ¢¢
dL
= (C¢¢¢
L
p – C¢¢
p)+
dT
T
v¢¢¢ - v¢¢
11.19 Show that for a van der Waals gas, the Joule–Thomson coefficient is given by
F I
H K
mj =
LM b
MN
g
2
OP
a f PQ
2
v 2 a v - b - RTbv
2
3
c p RTv - 2 a v - b
11.20 At 273.15 K the specific volumes of water and ice are 0.001 and 0.001091
m3/kg and the latent heat of fusion of ice is 334 kJ/kg. Determine the melting
point increase due to increase of pressure by 1 atm (101.325 kPa)
Ans. 0.00753 K
11.21 Calculate the latent heat of vaporization of steam formed by boiling water
under a pressure of 101.325 kPa. At a pressure near this, a rise of temperature
of 1 K causes an increase of vapour pressure of 3.62 kPa. Ans. 2257 kJ/kg
11.22 It is known that radiation exerts a pressure p = 1/3 u , where u is the energy per
unit volume.
(a) Show that
F
H
I
K
1
4
Ts - u dV
V
3
where s is the entropy per unit volume.
(b) Assuming u and s as functions of temperature only, show that
(i) u = As4/3
4
(ii) s =
aT 3
3
(iii) u = aT 4
where A is the constant of integration and a = 81/256 A3.
(c) Show that the average time radiation remains in a spherical enclosure of
radius r is given by
4r
t=
3c
where c is the speed of radiation.
(d) If EB is the energy emitted per unit area of spherical surface per unit time,
show that
du = Tds +
434
Engineering Thermodynamics
EB = s T 4
where s = ac/4 and T is the temperature of the surface.
11.23 Show that the inversion temperature of a van der Waals gas is given by
Ti = 2a/bR.
11.24 Show that:
FG ∂u IJ = T LM ∂b p / T g OP
H ∂v K
N ∂T Q
F ∂h I
L ∂bv / T g OP
(b) G J = T M
H ∂p K
N ∂T Q
2
(a)
T
v
2
T
v
11.25 Show that for a van der Waals gas at low pressure, a Joule–Thomson expansion from pressure p1 to p2 produces a temperature change which can be found
from the solution of
cp
T - Ti
p1 – p2 =
(T1 – T2) + Ti ln 2
T1 - Ti
b
where Ti is the inversion temperature.
11.26 Using the Redlick-Kwong equation of state, develop expressions for the
changes in entropy and internal energy of a gas in an isothermal process.
LM b g OP
MN b g PQ
L v bv + b g OP
3a
(u – u ) =
ln M
2 bT
NM v b v + b g QP
Ans. (s2 – s1)T = R ln
2
v 2 v1 + b
v2 - b
a
+
ln
3/ 2
v1 - b 2 bT
v1 v 2 + b
1 T
1/ 2
2
1
1
2
11.27 Find the change of entropy of a gas following Clausius equation of state at
constant temperature
v -b
p(v – b) = RT
Ans. R ln 2
v1 - b
11.28 (a) Show that for a van der Waals gas
b=
kT =
b
g
Rv2 v - b
3
b
g
2
g
2
RTv - 2a v - b
b
v2 v - b
b
g
2
RTv 3 - 2 a v - b
(b) What is the value of kT /b expressed in its simplest form?
(c) What do the above relations become when a = 0, b = 0 (ideal gas)?
11.29 (a) Show that
(i)
FG ∂u IJ = k ◊ c
H ∂p K b
v
v
(ii)
FG ∂u IJ = c − p
H ∂v K v
p
p
β
435
Thermodynamic Relations, Equilibrium and Stability
(b) Hence show that the slope of a reversible adiabatic process on p–v coordinates is
dp
g
=dv
kv
where k is the isothermal compressibility.
11.30 According to Berthelot, the temperature effect of the second virial coefficient
is given by
b
a
B¢(T) = - 3
T T
where a and b are constants. Show that according to Berthelot,
Tinv/TB = 3
11.31 The following expressions for the equation of state and the specific heat cp are
obeyed by a certain gas:
RT
v=
+ aT2 and cp = A + BT + Cp
p
where a, A, B, C are constants. Obtain an expression for (a) the Joule-Thomson
coefficient, and (b) the specific heat cv.
Ans. (a) mJ =
(b) cv = A + BT +
aT 2
A + BT + Cp
F
GH
CRT
v + a T2
–
R
v - aT 2
v - a T2
I
JK
11.32 Determine the maximum Joule–Thomson inversion temperature in terms of
the critical temperature Tc predicted by the
(a) van der Waals’ equations
(b) Redlich—Kwong equation
(c) Dieterici equation
Ans. (a) 6.75 Tc, (b) 5.34 Tc (c) 8Tc
11.33 From the virial form of the equation of state of a gas
RT
v=
+ RTB¢(T) + RTC¢(T)p +
p
show that the Joule-Thomson coefficient is
mJ =
LM
N
OP
Q
RT 2 dB¢ dC¢
+
p +…
cp dT dT
(b) For a van der Waals gas
bRT - a
B¢(T) =
R2 T 2
Show that the limiting value of m J at low pressure is
mJ =
11.34 Show that kT – ks =
1
cp
Tvb 2
cp
F 2 a - bI
H RT K
436
Engineering Thermodynamics
11.35 For a simple compressible system, show that
LM ∂u OP = T LM ∂p / T OP
N dv Q
N ∂T Q
L ∂h O
L ∂v / T OP
(b) M P = - T M
N ∂T Q
N dp Q
2
(a)
T
V
2
T
p
11.36 The liquid-vapour equilibrium curve for nitrogen over the range from the triple
point to the normal boiling point may be expressed by the relation:
log p = A – BT –
C
T
where p is the vapour pressure in mm Hg, T is the temperature in K, and
A = 7.782, B = 0.006265, and C = 341.6.
(a) Derive an expression for the enthalpy of vaporization hfg in terms of A, B,
C, T and vfg.
(b) Calculate hfg for nitrogen at 71.9 K with v fg = 11.530 cm3/gmol.
Ans. 5,790 J/gmol
11.37 For a gas obeying the van der Waals equation of state, show that:
(a) cp – cv =
R
b
1 - 2a v - b
g / RTv
2
3
LM ∂c OP = T LM ∂ p OP = 0 to prove that c is a function of temperature only
N dv Q N ∂T Q
L ∂c OP = - T LM ∂ v OP
(c) M
N dp Q
N ∂T Q
L 2av - 6abv OP
=RTM
MM e p - av + 2av j PP
N
Q
2
(b)
v
v
2
T
v
2
p
2
T
p
-4
3
2
-2
-3 3
to prove that cp for a van der Waals gas is not a function of temperature only.
(d) The relation between T and v is given by:
T(v – b)R/cv = constant
(e) The relation between p and v is given by:
LM p + a OP (v – b)
N vQ
2
1 + R/cv
= constant.
11.38 Nitrogen at a pressure of 250 atm and a temperature of 400 K expands reversibly and adiabatically in a turbine to an exhaust pressure of 5 atm. The flow
rate is 1 kg/s. Calculate the power output if nitrogen obeys the Redlich–Kwong
equation of state. For nitrogen at 1 atm take.
cp = 6.903 – 0.3753 ¥ 10–3 T + 1.930 ¥ 10–6 T2 – 6.861 ¥ 10–9 T3
where cp is in cal/gmol-K and T is in K.
For nitrogen, Tc = 126.2 K,
pc = 33.5 atm.
Ans. 272 kW
Thermodynamic Relations, Equilibrium and Stability
437
Hints: See Fig. P–11.34
h1 – h2 = (h1 – h4) + (h4 – h3) + (h3 – h2) and
s1 – s2 = 0 = (s1 – s4) + (s4 – s3) +(s3 – s 2)
a = 15.4 ¥ 106 atm/K1/2 cm6/(gmol)2, b = 26.8 cm3/gmol
By trial-and-error, v1 = 143 cm3/gmol, v 4 = 32,800 cm3/gmol
T2 = 124 K, h1 – h2 = 7.61 kJ/gmol.
12
Vapour Power Cycles
12.1
SIMPLE STEAM POWER CYCLE
A power cycle continuously converts heat (energy released by the burning of fuel)
into work (shaft work), in which a working fluid repeatedly performs a succession
of processes. The components of a simple vapour power plant are shown in
Fig. 12.1 (a). In the vapour power cycle the working fluid, which is water, undergoes
a change of phase. Figure 12.1 (b) gives the schematic of a simple steam power
plant working on the vapour power cycle. Heat is transferred to water in the boiler
from an external source (furnace, where fuel is continuously burnt) to raise steam,
the high pressure, high temperature steam leaving the boiler expands in the turbine
to produce shaft work, the steam leaving the turbine condenses into water in the
condenser (where cooling water circulates), rejecting heat, and then the water is
pumped back to the boiler. Figure 12.2 shows how a unit mass of the working fluid,
sometimes in the liquid phase and sometimes in the vapour phase, undergoes various
external heat and work interactions in executing a power cycle. Since the fluid is
undergoing a cyclic process, there will be no net change in its internal energy over
the cycle, and consequently the net energy transferred to the unit mass of the fluid
as heat during the cycle must equal the net energy transfer as work from the fluid.
Figure 12.3 shows the cyclic heat engine operating on the vapour power cycle,
where the working substance, water, follows along the B-T-C-P (Boiler-TurbineCondenser-Pump) path, interacting externally as shown, and converting net heat
input to net work output continuously. By the first law
 Qnet =  Wnet
Cycle
or
where
Cycle
Q1 – Q2 = WT – WP
Q1 = heat transferred to the working fluid (kJ/kg)
Q2 = heat rejected from the working fluid (kJ/kg)
WT = work transferred from the working fluid (kJ/kg)
WP = work transferred into the working fluid (kJ/kg)
439
Stack
Vapour Power Cycles
Combustion gases
to stack
Turbine
Boiler
Fuel
Condenser
Air
Warm water
Cooling
tower
+
Electric
generator
–
Pump
Feedwater
pump
Fig. 12.1
Cooled water
Makeup water
(a) Components of a Simple Vapour Power Plant
High pressure, high
temperature steam
WT
Furnace
(source)
Q1
Boiler
Turbine
Air
and
fuel
Combustion
products
Generator
River
(Sink)
T2
Condenser
Condensate
pump
High pressure water
Q2
Wp
Fig. 12.1 (b) Simple Steam Power Plant
The efficiency of the vapour power cycle would be given by
hcycle =
Wnet
W - WP Q1 - Q2
= T
=
Q1
Q1
Q1
Circulating
pump
440
Engineering Thermodynamics
Q1
WT
Q2
WP
4
1 kg
H2O
1 kg
H2O
1 kg
H2O
1 kg
H2O
State change
from 4 to 1
(in boiler)
State change
from 1 to 2
(in turbine)
State change
from 2 to 3
(in condenser)
State change
from 3 to 4
(in pump)
Fig. 12.2 One kg H2O Executing a Heat Engine Cycle
WT
Source
(Furnace)
T1
B
H
T
O
H2
Q1
2
1
4
2O
O
H2
P
3
WP
C
O
H2
Q2
Sink
(River or sea)
T2
Cyclic heat engine
Fig. 12.3 Cyclic Heat Engine With Water as the Working Fluid
=1–
12.2
Q2
Q1
(12.1)
RANKINE CYCLE
For each process in the vapour power cycle, it is possible to assume a hypothetical
or ideal process which represents the basic intended operation and involves no
extraneous effects. For the steam boiler, this would be a reversible constant pressure
heating process of water to form steam, for the turbine the ideal process would be a
reversible adiabatic expansion of steam, for the condenser it would be a reversible
constant pressure heat rejection as the steam condenses till it becomes saturated
liquid, and for the pump, the ideal process would be the reversible adiabatic
compression of this liquid ending at the initial pressure. When all these four
processes are ideal, the cycle is an ideal cycle, called a Rankine cycle. This is a
reversible cycle. Figure 12.4 shows the flow diagram of the Rankine cycle, and in
Fig. 12.5, the cycle has been plotted on the p-v, T-s, and h-s planes. The numbers on
the plots correspond to the numbers on the flow diagram. For any given pressure,
the steam approaching the turbine may be dry saturated (state 1) wet (state 1¢), or
superheated (state 1¢¢ ), but the fluid approaching the pump is, in each case, saturated
liquid (state 3). Steam expands reversibly and adiabatically in the turbine from
state 1 to state 2 (or 1¢ to 2¢, or 1¢¢ to 2¢¢), the steam leaving the turbine condenses
to water in the condenser reversibly at constant pressure from state 2 (or 2¢, or 2¢¢)
to state 3, the water at state 3 is then pumped to the boiler at state 4 reversibly and
adiabatically, and the water is heated in the boiler to form steam reversibly at
constant pressure from state 4 to state 1 (or 1¢ or 1¢¢).
441
Vapour Power Cycles
WT
Turbine
Boiler
Q1
Condenser
Cooling water
3
Q2
4
Pump
Fig. 12.4
WP
A Simple Steam Plant
For a simple nuclear power plant, the boiler is replaced by a nuclear reactor (Fig.
12.4 a) where heat released by nuclear fission is utilized in the generation of steam.
Other features of the plant are similar to a conventional steam plant (Fig. 12.5).
H2O
1
Q
WT
Steam
turbine
Nuclear
reactor
2
4
Condenser
3
C.W.
Q2
WP
Pump
Fig. 12.4 (a) A Simple Nuclear Power Plant
For purposes of analysis the Rankine cycle is assumed to be carried out in a
steady flow operation. Applying the steady flow energy equation to each of the
processes on the basis of unit mass of fluid, and neglecting changes in kinetic and
potential energy, the work and heat quantities can be evaluated in terms of the
properties of the fluid.
442
Engineering Thermodynamics
1¢¢
p
1
1¢¢
T1
4
p2
3
p1 1¢
T
1
4 1¢
p1
2¢ 2
p2
3
2¢¢
2¢ 2
2¢¢
s
v
(a)
(b)
h
1¢¢
1
1¢
p1
p2
4
3
2¢ 2
2¢¢
s
(c)
Fig. 12.5 Rankine Cycle on p-v, T-s and h-s Diagrams
For 1 kg fluid
The S.F.E.E. for the boiler (control volume) gives
h4 + Q1 = h1
\
Q1 = h1 – h4
The S.F.E.E. for the turbine as the control volume gives
h1 = WT + h2
(12.2)
\
WT = h1 – h2
Similarly, the S.F.E.E. for the condenser is
h2 = Q2 + h3
(12.3)
\
Q2 = h2 – h3
and the S.F.E.E. for the pump gives
h3 + Wp = h4
(12.4)
\
Wp = h4 – h3
The efficiency of the Rankine cycle is then given by
(12.5)
h=
b
g b
h1 - h2 - h4 - h3
Wnet
W - WP
= T
=
Q1
Q1
h1 - h4
g
(12.6)
443
Vapour Power Cycles
The pump handles liquid water which is incompressible, i.e. its density or specific
volume undergoes little change with an increase in pressure. For reversible adiabatic
compression, by the use of the general property relation
Tds = dh – vdp;
and
ds = 0
dh = vdp
Since change in specific volume is negligible
Dh = v Dp
or
h4 – h3 = v3 ( p1 – p2 )
3
If v is in m /kg and p is in bar
h4 – h3 = v3 ( p1 – p2) ¥ 105 J/kg
(12.7)
Usually, the pump work is quite small compared to the turbine work and is
sometimes neglected. Then h4 = h3, and the cycle efficiency approximately becomes
h@
h1 - h2
h1 - h4
1
T
The efficiency of the Rankine
cycle is presented graphically in the
Wnet
T-s plot in Fig. 12.6. Thus Q1 is proportional to area 1564, Q2 is
4
2
proportional to area 2563, and Wnet
3
Q2
(= Q1 – Q2) is proportio-nal to area
6
5
1234 enclosed by the cycle.
s
The capacity of a steam plant is Fig. 12.6 Q , W and Q are
1
net
2
often expressed in terms of steam
Proportional to Areas
rate, which is defined as the rate of
steam flow (kg/h) required to produce unit shaft output (1 kW). Therefore
Steam rate =
=
1
kg 1 kJ/s
◊
WT - WP kJ 1 kW
1
3600
kg
kJ
=
WT - WP kWs WT - WP kWh
(12.8)
The cycle efficiency is sometimes expressed alternatively as heat rate which is
the rate of heat input (Q1) required to produce unit work output (1 kW)
Heat rate =
3600 Q1
3600 kJ
=
h cycle kWh
WT - WP
(12.9)
−4
From the equation Wrev = −
z
v dp, it is obvious that the reversible steady-flow
3
work is closely associated with the specific volume of fluid flowing through the
device. The larger the specific volume, the larger the reversible work produced or
444
Engineering Thermodynamics
consumed by the steady-flow device. Therefore, every effort should be made to
keep the specific volume of a fluid as small as possible during a compression
process to minimize the work input and as large as possible, during an expansion
process to maximize the work output.
In steam or gas power plants (Chapter 13), the pressure rise in the pump or
compressor is equal to the pressure drop in the turbine if we neglect the pressure
losses in various other components. In steam power plants, the pump handles liquid,
which has a very small specific volume, and the turbine handles vapour, whose
specific volume is many times larger. Therefore, the work output of the turbine is
much larger than the work input to the pump. This is one of the reasons for the
overwhelming popularity of steam power plants in electric power generation.
If we were to compress the steam exiting the turbine back to the turbine inlet
pressure before cooling it first in the condenser in order to “save” the heat rejected,
we would have to supply all the work produced by the turbine back to the
compressor. In reality, the required work input would be still greater than the work
output of the turbine because of the irreversibilities present in both processes (see
Example 12.1).
12.3 ACTUAL VAPOUR CYCLE PROCESSES
The processes of an actual cycle differ from those of the ideal cycle. In the actual
cycle conditions might be as indicated in Figs. 12.7 and 12.8, showing the various
losses. The thermal efficiency of the cycle is
Qloss
p1¢ > p1
1¢
1
Qloss
Turbine
WT < (h1 – h2)
< (h1 – h2s)
Boiler
2
4¢
Q2
p4¢ < p4 Condenser
p3 < p 2
3
Pump
Qloss
4
Qloss
Wp > (h4 – h3) > (h4s – h3)
Fig. 12.7 Various Losses in a Steam Plant
445
Vapour Power Cycles
hth =
Wnet
Q1
where the work and heat quantities are the measured values for the actual cycle,
which are different from the corresponding quantities of the ideal cycle.
T
p4
p4
1
p1
p1
1
1
4
4s
4¢
3
p2
p3
2¢ 2s 2
s
Fig. 12.8
12.3.1
Various Losses on T-s Plot
Piping Losses
Pressure drop due to friction and heat loss to the surroundings are the most
important piping losses. States 1¢ and 1 (Fig. 12.8) represent the states of the steam
leaving the boiler and entering the turbine respectively, 1¢ – 1¢¢ represents the
frictional losses, and 1¢¢ – 1 shows the constant pressure heat loss to the
surroundings. Both the pressure drop and heat transfer reduce the availability of
steam entering the turbine.
A similar loss is the pressure drop in the boiler and also in the pipeline from the
pump to the boiler. Due to this pressure drop, the water entering the boiler must be
pumped to a much higher pressure than the desired steam pressure leaving the boiler,
and this requires additional pump work.
12.3.2
Turbine Losses
The losses in the turbine are those associated with frictional effects and heat loss to
the surroundings. The steady flow energy equation for the turbine in Fig. 12.7 gives
h1 = h2 + WT + Qloss
\
WT = h1 – h2 – Qloss
(12.10)
For the reversible adiabatic expansion, the path will be 1 – 2s. For an ordinary real
turbine the heat loss is small, and WT is h1 – h2, with Q2 equal to zero. Since actual
turbine work is less than the reversible ideal work output, h2 is greater than h2s.
However, if there is heat loss to the surroundings, h2 will decrease, accompanied by
a decrease in entropy. If the heat loss is large, the end state of steam from the turbine
446
Engineering Thermodynamics
may be 2¢. It may so happen that the entropy increase due to frictional effects just
balances the entropy decrease due to heat loss, with the result that the initial and
final entropies of steam in the expansion process are equal, but the expansion is
neither adiabatic nor reversible. Except for very small turbines, heat loss from
turbines is generally negligible. The isentropic efficiency of the turbine is defined
as follows:
hT =
WT
h - h2
= 1
h1 - h2 s
h1 - h2 s
(12.11)
where WT is the actual turbine work, and (h1 – h2s) is the isentropic enthalpy drop in
the turbine (i.e. the ideal output).
12.3.3
Pump Losses
The losses in the pump are similar to those of the turbine, and are primarily due to
the irreversibilities associated with fluid friction. Heat transfer is usually negligible.
The pump efficiency is defined as
hP =
h4s - h3
h - h3
= 4s
WP
h4 - h3
(12.12)
where WP is the actual pump work.
12.3.4
Condenser Losses
The losses in the condenser are usually small. These include the loss of pressure
and the cooling of condensate below the saturation temperature.
12.4 COMPARISON OF RANKINE AND CARNOT CYCLES
Although the Carnot cycle has the maximum possible efficiency for the given limits
of temperature, it is not suitable in steam power plants. Figure 12.9 shows the
Rankine and Carnot cycles on the T–s diagram. The reversible adiabatic expansion
in the turbine, the constant temperature heat rejection in the condenser, and the
reversible adiabatic compression in the pump, are similar characteristic features of
both the Rankine and Carnot cycles. But whereas the heat addition process in the
Rankine cycle is reversible and at constant pressure, in the Carnot cycle it is
reversible and isothermal. In Figs. 12.9 (a) and 12.9 (c), Q2 is the same in both the
cycles, but since Q1 is more, hCarnot is greater than hRankine . The two Carnot cycles
in Fig. 12.9(a) and 12.9 (b) have the same thermal efficiency. Therefore, in Fig.
12.9 (b) also, hCarnot > hRankine . But the Carnot cycle cannot be realized in practice
because the pump work [in all the three cycles (a), (b), and (c)] is very large.
Whereas in (a) and (c) it is impossible to add heat at infinite pressures and at
constant temperature from state 4c to state 1, in (b), it is difficult to control the
quality at 3c, so that insentropic compression leads to saturated liquid state.
447
1
4C
1
T
T
Vapour Power Cycles
4C
4R
4R
3
3
2
3C
2s
s
s
(a)
(b)
4C
T
1
4R
3
2s
(c)
Fig. 12.9 Comparison of Carnot and Rankine Cycles
12.5 MEAN TEMPERATURE OF HEAT ADDITION
Q1 = h1 – h4s = Tm1 (s1 – s4s)
\ Tm1 = Mean temperature of
heat addition
=
T
In the Rankine cycle, heat is added reversibly at a constant pressure, but at infinite
temperatures. If Tm1 is the mean temperature of heat addition, as shown in
Fig. 12.10, so that the area under
1
4s and 1 is equal to the area under
5 and 6, then heat added
5
Tm1
6
4s
h1 - h4 s
s1 - s4s
3
Since Q2 = Heat rejected = h2s –
h3
= T2 (s1 – s4s)
hRankine = 1 –
2s
T2
s
Fig. 12.10 Mean Temperature of Heat Addition
b
b
g
g
T2 s1 - s4 s
Q2
=1–
Tm1 s1 - s4 s
Q1
448
\
hRankine = 1 –
Engineering Thermodynamics
T2
Tm1
(12.13)
where T2 is the temperature of heat rejection. The lower is the T2 for a given Tm1 ,
the higher will be the efficiency of the Rankine cycle. But the lowest practicable
temperature of heat rejection is the temperature of the surroundings (T0). This being
fixed,
hRankine = f(Tm1) only
(12.14)
The higher the mean temperature of heat addition, the higher will be the cycle
efficiency.
1¢
T
T
The effect of increasing the
1
initial temperature at constant
pressure on cycle efficiency is
shown in Fig. 12.11. When the
initial state changes from 1 to
4s
1¢, Tm1 between 1 and 1¢ is
higher than Tm1 between 4s and
2s 2¢s
3
1. So an increase in the superheat at constant pressure
s
increases the mean temperature
of heat addition and hence the
Fig. 12.11 Effect of Superheat on Mean
cycle efficiency.
Temperature of Heat Addition
The maximum temperature
of steam that can be used is fixed from metallurgical considerations (i.e. the
materials used for the manufacture of the components which are subjected to highpressure, high-temperature steam like the superheaters, valves, pipelines, inlet
stages of turbine, etc.).
When the maximum
5 1
Tmax
temperature is fixed, as
the operating steam
p2
pressure at which heat is
added in the boiler
increases from p1 to p2
p1
(Fig. 12.12), the mean
temperature of heat
7s
addition increases, since
4s
T2
Tm1 between 7s and 5 is
6s 2s
3
higher than that between
x6s x2s
4s and 1. But when the
s
turbine inlet pressure
increases from p1 to p2,
Fig. 12.12 Effect of Increase of Pressure
the ideal expansion line
on Rankine Cycle
shifts to the left and the
moisture content at the turbine exhaust increases (because x6s < x2s). If the moisture
content of steam in the later stages of the turbine is higher, the entrained water
Vapour Power Cycles
449
T
particles along with the vapour coming out from the nozzles with high velocity
strike the blades and erode their surfaces, as a result of which the longevity of the
blades decreases. From a consideration of the erosion of blades in the later stages
of the turbine, the maximum moisture content at the turbine exhaust is not allowed
to exceed 15%, or the quality to fall below 85%. It is desirable that most of the
turbine expansion should take place in the single phase or vapour region.
Therefore, with the maximum steam temperature at the turbine inlet, the
minimum temperature of
heat rejection, and the
1
Tmax
minimum quality of steam
at the turbine exhaust being
fixed, the maximum steam
( p1)max
pressure at the turbine inlet
also gets fixed (Fig. 12.13).
The vertical line drawn
from 2s, fixed by T2 and x2s,
4s
p2
intersects the Tmax line,
T2
3
fixed by material, at 1,
2s
x2s = 0.85
which gives the maximum
s
steam pressure at the turbine
inlet. The irrever sibility in
Fig. 12.13
Fixing of Exhaust Quality, Maximum
the expansion process has,
Temperature and Maximum Pressure in
however,
not
been
Rankine Cycle
considered.
12.6 REHEAT CYCLE
If a steam pressure higher than (p1)max (Fig. 12.13) is used, in order to limit the
quality to 0.85, at the turbine exhaust, reheat has to be adopted. In that case all the
steam after partial expansion in the turbine is brought back to the boiler, reheated
by combustion gases and then fed back to the turbine for further expansion. The
flow, T-s, and h-s diagrams for the ideal Rankine cycle with reheat are shown in
Fig. 12.14. In the reheat cycle the expansion of steam from the initial state 1 to the
condenser pressure is carried out in two or more steps, depending upon the number
of reheats used. In the first step, steam expands in the high pressure (H.P.) turbine
from the initial state to approximately the saturated vapour line (process 1–2s in
Fig. 12.14). The steam is then resuperheated (or reheated) at constant pressure in
the boiler (process 2s–3) and the remaining expansion (process 3–4s) is carried out
in the low pressure (L.P.) turbine. In the case of use of two reheats, steam is
resuperheated twice at two different constant pressures. To protect the reheat tubes,
steam is not allowed to expand deep into the two-phase region before it is taken for
reheating, because in that case the moisture particles in steam while evaporating
would leave behind solid deposits in the form of scale which is difficult to remove.
Also, a low reheat pressure may bring down Tm1 and hence, cycle efficiency. Again,
a high reheat pressure increases the moisture content at turbine exhaust. Thus, the
reheat pressure is optimized. The optimum reheat pressure for most of the modern
power plants is about 0.2 to 0.25 of the initial steam pressure. For the cycle in
Fig. 12.14, for 1 kg of steam
450
Engineering Thermodynamics
p1, t1
WT
L.P.
Turbine
H.P.
Turbine
Boiler
Reheater
Q1
Generator
p3
p3, t3
p2
Condenser
Q2
Pump
WP
(a)
1
3
t 1 = t3
T
p1
2s
p3
6s
p2
5
4¢s
x ¢4s 4s x 4s
s
(b)
3
h
1
Critical
point
p1
2s
p3
p2
6s
t 1 = t3
4s
4¢s
5
x4s
¢
s
(c)
Fig. 12.14
Reheat Cycle
x 4s
451
Vapour Power Cycles
Q1 = h1 – h6s + h3 – h2s
Q2 = h4s – h5
WT = h1 – h2s + h3 – h4s
WP = h6s – h5
h=
Steam rate =
b
g b
h1 - h2 s + h3 - h4 s - h6 s - h5
WT - WP
=
h1 - h6 s + h3 - h2 s
Q1
b
3600
kg/kWh
h1 - h2 s + h3 - h4 s - h6s - h5
g b
g
g
(12.15)
(12.16)
where enthalpy is in kJ/kg.
Since higher pressures are used in a reheat cycle, pump work may be appreciable.
Had the high pressure p1 been used without reheat, the ideal Rankine cycle would
have been 1 – 4¢s – 5 – 6s. With the use of reheat, the area 2s – 3 – 4s – 4¢s has been
added to the basic cycle. It is obvious that net work output of the plant increases
with reheat, because (h3 – h4s) is greater than (h2s – h4¢s), and hence the steam rate
decreases. Whether the cycle efficiency improves with reheat depends upon whether
the mean temperature of heat addition in process 2s–3 is higher than the mean
temperature of heat addition in process 6s – 1. In practice, the use of reheat only
gives a small increase in cycle efficiency, but it increases the net work output by
making possible the use of higher pressures, keeping the quality of steam at turbine
exhaust within a permissible limit. The quality improves from x¢4 s to x4s by the use
of reheat.
By increasing the number of reheats, still higher steam pressures could be used,
but the mechanical stresses increase at a higher proportion than the increase in
pressure, because of the prevailing high temperature. The cost and fabrication
difficulties will also increase. In that way, the maximum steam pressure gets fixed,
and more than two reheats have not yet been used so far.
In Fig. 12.14, only ideal processes have been considered. The irreversibilities in
the expansion and compression processes have been considered in the example
given later.
12.7 IDEAL REGENERATIVE CYCLE
In order to increase the mean temperature of heat addition (Tm1), attention was so
far confined to increasing the amount of heat supplied at high temperatures, such as
increasing superheat, using higher pressure and temperature of steam, and using
reheat. The mean temperature of heat addition can also be increased by decreasing
the amount of heat added at low temperatures. In a saturated steam Rankine cycle
(Fig. 12.15), a considerable part of the total heat supplied is in the liquid phase
when heating up water from 4 to 4¢, at a temperature lower than T1, the maximum
temperature of the cycle. For maximum efficiency, all heat should be supplied at T1,
and feedwater should enter the boiler at state 4¢. This may be accomplished in what
452
Engineering Thermodynamics
is known as an ideal regenerative cycle, the flow diagram of which is shown in Fig.
12.16 and the corresponding T-s diagram in Fig. 12.17.
4¢
T
1
T1
4
3
2
s
Fig. 12.15
Simple Rankine Cycle
The unique feature of the ideal regenerative cycle is that the condensate, after
leaving the pump circulates around the turbine casing, counterflow to the direction
of vapour flow in the turbine (Fig. 12.16). Thus it is possible to transfer heat from
the vapour as it flows through the turbine to the liquid flowing around the turbine.
Let us assume that this is a reversible heat transfer, i.e. at each point the temperature
of the vapour is only infinitesimally higher than the temperature of the liquid. The
process 1–2¢ (Fig. 12.17) thus represents reversible expansion of steam in the
turbine with reversible heat rejection. For any small step in the process of heating
the water,
DT (water) = – DT (steam)
Ds (water) = – D s (steam)
and
WT
Q1
Q2
WP
Fig. 12.16 Ideal Regenerative Cycle—Basic Scheme
453
T
Vapour Power Cycles
(Ds)w
4¢
1
(Ds)st
T1
DT
4
3
5
2¢
a
b
2
c
T2
d
s
Fig. 12.17 Ideal Regenerative Cycle on T-s Plot
Then the slopes of lines 1–2¢ and 4¢–3 (Fig. 12.17) will be identical at every
temperature and the lines will be identical in contour. Areas 4–4¢–b–a–4 and
2¢–1–d–c-2¢ are not only equal but congruous. Therefore, all the heat added from
an external source (Q1) is at the constant temperature T1, and all the heat rejected
(Q2) is at the constant temperature T2, both being reversible. Then
Q1 = h1 – h4¢ = T1(s1 – s4¢ )
Q2 = h2¢ – h3 = T2 (s2¢ – s3)
Since
s4¢ – s3 = s1 – s2¢
or
s1 – s4¢ = s2¢ – s3
\
h=1–
Q2
T
=1– 2
Q1
T1
The efficiency of the ideal regenerative cycle is thus equal to the Carnot cycle
efficiency.
Writing the steady flow energy equation for the turbine
h1 – WT – h 2¢ + h4 – h 4¢ = 0
or
WT = (h1 – h 2¢ ) – (h 4¢ – h4)
(12.17)
The pump work remains the same as in the Rankine cycle, i.e.
WP = h4 – h3
The net work output of the ideal regenerative cycle is thus less, and hence its steam
rate will be more, although it is more efficient, when compared with the Rankine
cycle. However, the cycle is not practicable for the following reasons:
(a) Reversible heat transfer cannot be obtained in finite time.
(b) Heat exchanger in the turbine is mechanically impracticable.
(c) The moisture content of the steam in the turbine will be high.
454
12.8
Engineering Thermodynamics
REGENERATIVE CYCLE
In the practical regenerative cycle, the feedwater enters the boiler at a temperature
between 4 and 4¢ (Fig. 12.17), and it is heated by steam extracted from intermediate
stages of the turbine. The flow diagram of the regenerative cycle with saturated
steam at the inlet to the turbine, and the corresponding T-s diagram are shown in
Figs. 12.18 and 12.19 respectively. For every kg of steam entering the turbine, let
m1 kg steam be extracted from an intermediate stage of the turbine where the
pressure is p2, and it is used to heat up feedwater [(1 – m1) kg at state 8] by mixing
in heater 1. The remaining (1 – m1) kg of steam then expands in the turbine from
pressure p2 (state 2) to pressure p3 (state 3) when m2 kg of steam is extracted for
heating feed water in heater 2. So (1 – m1 – m2) kg of steam then expands in the
remaining stages of the turbine to pressure p4, gets condensed into water in the
condenser, and then pumped to heater 2, where it mixes with m2 kg of steam
extracted at pressure p3. Then (1 – m1) kg of water is pumped to heater 1 where it
mixes with m1 kg of steam extracted at pressure p2. The resulting 1 kg of steam is
then pumped to the boiler where heat from an external source is supplied. Heaters 1
and 2 thus operate at pressures p2 and p3 respectively. The amounts of steam m1 and
m2 extracted from the turbine are such that at the exit from each of the heaters, the
state is saturated liquid at the respective pressures. The heat and work transfer
quantities of the cycle are as follows:
1 kg
p1, tsat
WT
Turbine
Boiler
Generator
p4
Q1
m1 kg
m2 kg
p2
p3
(1 - m1 - m2) kg
Condenser
C.V.
C.V.
1-m1
Heater-1
9
8
Q2
Heater-2
7
6
5
Pump 1
10
WP3
Pump 3
Fig. 12.18
WP2
W P1
Pump 2
Regenerative Cycle Flow Diagram with two Feedwater Heaters
Vapour Power Cycles
WT = 1 (h1 – h2) + (1 – m1) (h2 – h3) + (1 – m1 – m2) (h3 – h4) kJ/kg
455
(12.18)
WP = WP1 + WP2 + WP3
= (1 – m1 – m2) (h6 – h5) + (1 – m1) (h8 – h7) + 1 (h10 – h9) kJ/kg
(12.19)
Q1 = 1 (h1 – h10 ) kJ/kg
(12.20)
Q2 = (1 – m1 – m2) (h4 – h5) kJ/kg
(12.21)
Cycle efficiency, h =
Q1 - Q2
W - WP
= T
Q1
Q1
Steam rate =
3600
kg/kW h
WT - WP
In the Rankine cycle operating at the given pressures, p1 and p4, the heat addition
would have been from state 6 to state 1. By using two stages of regenerative
feedwater heating, feedwater enters the boiler at state 10, instead of state 6, and
heat addition is, therefore, from state 10 to state 1. Therefore
and
(Tm1)with regeneration =
h1 - h10
s1 - s10
(12.22)
(Tm1)without regeneration =
h1 - h6
s1 - s6
(12.23)
Since
(Tm1)with regeneration > (Tm1)without regeneration
the efficiency of the regenerative cycle will be higher than that of the Rankine cycle.
The energy balance for heater 2 gives
m1h2 + (1 – m1) h8 = 1h9
∵
m1 =
h9 - h8
h2 - h8
(12.24)
The energy balance for heater 1 gives
m2h3 + (1 – m1 – m2) h6 = (1 – m1) h7
or
m2 = (1 – m1)
h7 - h6
h3 - h6
(12.25)
From Eqs. (12.24) and (12.25), m1 and m2 can be evaluated. Equations (12.24) and
(12.25) can also be written alternatively as
(1 – m1) (h9 – h8) = m1 (h2 – h9)
(1 – m1 – m2 ) (h7 – h6) = m2 (h3 – h7)
Energy gain of feedwater = Energy given off by vapour in condensation
Heaters have been assumed to be adequately insulated, and there is no heat gain
from, or heat loss to, the surroundings.
456
Engineering Thermodynamics
1
1 kg
p1
1 kg
m1 kg
T
10
9
p2
m2 kg
2
p3
7
(1 - m1 - m2) kg
3
8
6
(1 - m1) kg
(1 - m1 - m2) kg
p4
5
4
s
(a)
1
1 kg
T
10
p1
Loss in work
output
2¢
9
8
7
1 kg
6
5
p4
p2
p3
1 kg
3¢¢
2
3¢
3
1 kg
4¢
4
s
(b)
Fig. 12.19
(a)Regenerative Cycle on T-s Plot with Decreasing Mass of Fluid
(b)Regenerative Cycle on T-s Plot for Unit Mass of Fluid
Path 1–2–3–4 in Fig. 12.19 represents the states of a decreasing mass of fluid.
For 1 kg of steam, the states would be represented by the path 1–2¢–3¢–4¢. From
Eq. (12.18),
WT = (h1 – h2) + (1 – m1) (h2 – h3) + (1 – m1 – m2) (h3 – h4 )
= (h1 – h2) + (h 2¢ – h 3¢) + (h 3¢ – h 4¢)
where
(1 – m1) (h2 – h3) = 1 (h 2¢ – h 3¢)
(12.26)
(12.27)
(1 – m1 – m2) (h3 – h4) = 1 (h3¢¢ – h4¢)
(12.28)
The cycle 1–2–2¢–3¢–3¢¢–4¢–5–6–7–8–9–10–1 represents 1 kg of working fluid.
The heat released by steam condensing from 2 to 2¢ is utilized in heating up the
water from 8 to 9.
∵
1(h2 – h 2¢) = (h9 – h8)
(12.29)
Similarly
1(h 3¢ – h 3¢¢) = 1(h7 – h6)
(12.30)
457
Vapour Power Cycles
From Eqs. (12.26), (12.29) and (12.30)
WT = (h1 – h 4¢) – (h2 – h 2¢) – (h 3¢ – h 3¢¢)
T
= (h1 – h 4¢) – (h9 – h8) – (h7 – h6)
(12.31)
The similarity of Eqs. (12.17) and (12.31) can be noticed. It is seen that the stepped
cycle 1–2¢–3¢–4¢–5–6–7–8–9–10 approximates the ideal regenerative cycle in
Fig. 12.17, and that a greater number of stages would give a closer approximation
(Fig. 12.20). Thus the heating of feedwater by steam ‘bled’ from the turbine, known
as regeneration, carnotizes the Rankine cycle.
s
Fig. 12.20
Regenerative Cycle with many Stages of Feedwater Heating
The heat rejected Q2 in the cycle decreases from (h4 – h5) to (h 4¢ – h5). There is
also loss in work output by the amount (Area under 2–2¢ + Area under 3¢–3¢¢ – Area
under 4–4¢), as shown by the hatched area in Fig. 12.19 (b). So the steam rate
increases by regeneration, i.e. more steam has to circulate per hour to produce unit
shaft output.
The enthalpy-entropy diagram of a regenerative cycle is shown in Fig. 12.21.
p1
Critical point
p2
1
p3
g
h
1k
m1
10
8
6
kg
p4
2
kg 3¢¢
3¢
9
7
5
m2
2¢
g
- m2) k
(1 - m1
4¢
3
4
s
Fig. 12.21
Regenerative Cycle on h-s Diagram
12.9 REHEAT-REGENERATIVE CYCLE
The reheating of steam is adopted when the vaporization pressure is high. The effect
of reheat alone on the thermal efficiency of the cycle is very small. Regeneration or
the heating up of feedwater by steam extracted from the turbine has a marked effect
458
Engineering Thermodynamics
on cycle efficiency. A modern steam power plant is equipped with both. Figures
12.22 and 12.23 give the flow and T-s diagrams of a steam plant with reheat and
three stages of feedwater heating. Here
WT = (h1 – h2) + (1 – m1) (h2 – h3) + (1 – m1) (h4 – h5)
+ (1 – m1 – m2) (h5 – h6) + (1 – m1 – m2 – m3) (h6 – h7) kJ/kg
WP = (1 – m1 – m2 – m3) (h9 – h8) + (1 – m1 – m2 ) (h11 – h10)
+ (1 – m1) (h13 – h12 ) + 1(h15 – h14 ) kJ/kg
Q1 = (h1 – h15) + (1 – m1) (h4 – h3) kJ/kg
and
Q2 = (1 – m1 – m2 – m3) (h7 – h8) kJ/kg
1 kg
W1
Boiler
L.P.
Turbine
H.P.
Turbine
2
(1 - m1) kg
3
4
Q1
5
6
7
m1 kg
Reheater
(1 - m1 - m2 - m3) kg
m2 kg
Heater
m3 kg
3
Heater
2
Heater
1
Q2
14
15
13 12
WP4
10
11
WP3
9
8
WP2
WP1
Reheat-Regenerative Cycle Flow Diagram
Fig. 12.22
1
4
(1 - m1) kg
2
T
1 kg
15
5
m1 kg
14
3
(1 - m1 - m2) kg
m2 kg
13
12
6
m3 kg
11
10
9
(1– m1 – m2 – m3) kg
8
7
s
Fig. 12.23
T-s Diagram of Reheat-Regenerative Cycle
459
Vapour Power Cycles
The energy balance of heaters 1, 2, and 3 give
m1h2 + (1 – m1)h13 = 1 ¥ h14
m2h5 + (1 – m1 – m2)h11 = (1 – m1)h12
m3h6 + (1 – m1 – m2 – m3)h9 = (1 – m1 – m2)h10
from which m1, m2, and m3 can be evaluated.
12.10
FEEDWATER HEATERS
Feedwater heaters are of two types, viz. open heaters and closed heaters. In an open
or contact-type heater, the extracted or bled steam is allowed to mix with feedwater,
and both leave the heater at a common temperature, as shown in Figs. 12.18 and
12.22. In a closed heater, the fluids are kept separate, and not allowed to mix
together (Fig. 12.24). The feedwater flows through the tubes in the heater and the
extracted steam condenses on the outside of the tubes in the shell. The heat released
by condensation is transferred to the feedwater through the walls of the tubes. The
condensate (saturated water at the steam extraction pressure), sometimes called the
heater-drip, then passes through a trap into the next lower pressure heater. This, to
some extent, reduces the steam required by that heater. The trap passes only liquid
and no vapour. The drip from the lowest pressure heater could similarly be trapped
to the condenser, but this would be throwing away energy to the condenser cooling
water. To avoid this waste, a drip pump feeds the drip directly into the feedwater
stream.
1 kg
d
Turbine
Boiler
(1- m1 - m2) kg
e
i
g
m1 kg
m2 kg
Closed
heater
m1
Closed
heater
n
l
Condensate
pump
o
Trap
Trap
f
j
(m1 + m2) k
kg
h
Drip pump
m
Fig. 12.24
Regenerative Cycle flow Diagram with Closed Feedwater Heaters
460
Engineering Thermodynamics
Figure 12.25 shows the T-s plot corresponding to the flow diagram in
Fig. 12.24. The temperature of the feedwater (at ‘l ’ or ‘O’) leaving a particular
heater is always less than the saturation temperature at the steam extraction pressure
(at e or g), the difference being known as the terminal temperature difference of the
heater.
d
T
1 kg
o
m
n
l
k
m1 kg
f
e
(1- m1) kg
m2 kg
g
h
(1- m1 - m2) kg
j
i
s
Fig. 12.25 T-s diagram of Regenerative Cycle with Closed Feedwater Heaters
The advantages of the open heater are simplicity, lower cost, and high heat
transfer capacity. The disadvantage is the necessity of a pump at each heater to
handle the large feedwater stream.
A closed heater system requires only a single pump for the main feedwater stream
regardless of the number of heaters. The drip pump, if used, is relatively small.
Closed heaters are costly and may not give as high a feedwater temperature as do
open heaters. In most steam power plants, closed heaters are favoured, but at least
one open heater is used, primarily for the purpose of feedwater deaeration. The
open heater in such a system is called the deaerator.
An example of a power plant along with the T-s diagram is shown in Fig. 12.26.
The higher the number of heaters used, the higher will be the cycle efficiency. If
n heaters are used, the greatest gain in efficiency occurs when the overall
temperature rise is about n/(n + 1) times the difference between the condenser and
boiler saturation temperatures. (See Analysis of Engineering Cycles by R.W.
Haywood, Pergamon Press, 1973).
n=0
n = 1,
n = 2,
n = 3,
n = 4,
(Dt)fw = 0
1
2
2
(Dt)fw =
3
3
(Dt)fw =
4
4
(Dt)fw =
2
(Dt)fw =
(Dt)0
(Dt)0
(Dt)0
(Dt)0
1 (Dt)
0
6
1 (Dt)
Gain =
0
12
Gain =
Gain =
1 (Dt)
0
20
Qin
21
15
Closed
heater
Steam
generator
16
14
1
2
18
13
Fig. 12.26
17
Wp2
Closed
heater
4
5
(a)
Deaerating
open
heater
6
7
19
11
(a) Example of a Power Plant Layout
Main boiler
feed pump
12
3
Wp1
Closed
heater
Wt
10
20
Condensate
pump
9
Condenser
8
Qout
Vapour Power Cycles
461
462
Engineering Thermodynamics
1
1 kg
15
1 – m1 – m2
2
m1
21
6
3
m2
14
T
5
17 16
13
1 – m1 – m2 – m3
4
m3
12
7
18
m4
11
19
10
1 – m1 – m2 – m3 – m4
1 – m1 – m2 – m3 – m4
9
20
8
S
(b) T-s Diagram
Fig. 12.26
hcycle
If (Dt)0 = tboiler sat – tcond and (DT)fw = temperature rise of feedwaters, it is seen
that since the cycle efficiency is proportional to (Dt)fw the efficiency gain follows
the law of diminishing return with the increase in the number of heaters. The
greatest increment in efficiency occurs by the use of the first heater. (Fig. 12.27)
0
1
2
3
4
5
Number of heaters, n
Fig. 12.27 Effect of the use of Number of Heaters on Cycle Efficiency
463
Vapour Power Cycles
The increments for each additional heater thereafter successively diminish. The
number of heaters is fixed up by the energy balance of the whole plant when it is
found that the cost of adding another does not justify the saving in Q1 or the marginal
increase in cycle efficiency. An increase in feedwater temperature may, in some
cases, cause a reduction in boiler efficiency. So the number of heaters gets
optimized. Five points of extraction are often used in practice. Some cycles use as
many as nine.
12.11 EXERGY ANALYSIS OF VAPOUR POWER CYCLES
Let the heating for steam generation in the boiler unit be provided by a stream of
hot gases produced by burning of a fuel (Fig. 12.28 (a). The distribution of input
energy is shown in the Sankey diagram (12.28 (b)) which indicates that only about
30% of the input energy to the simple ideal plant is converted to shaft work and
about 60% is lost to the condenser. The exergy analysis, however, gives a different
distribution as discussed below.
Assuming that the hot gases are at atmospheric pressure, the exergy input is
LM
N
OP
Q
LT
T O
= w c T M - 1 - ln P
T
T
N
Q
af1 = wg cpg Ti - T0 - T0 ln
g
pg
Ti
T0
i
i
0
0
0
Ti
s
ase
1
eg
Flu
T
Energy lost with
waste gases
(~ 10%)
Energy
input (Q1)
Te
4
3
2
s
Wnet (~ 30%)
(a)
Q2 (~ 60%)
(b)
Exhaust flue
gases (~ 3%)
Exergy
input
Steam
Wnet
Condenser
generator (~ 47%)
(~ 4%)
(~ 46%)
(c)
Fig. 12.28
(a) T-s Diagram, (b) Sankey Diagram, (c) Grassman Diagram
464
Engineering Thermodynamics
Similarly, the exergy loss rate with the exhaust stream is:
af2 = wg cpg T0
LM T - 1 - ln T OP
T Q
NT
e
e
0
0
Net exergy input rate in the steam generation process:
ai1 = af1 – af2
The exergy utilization rate in the steam generation is:
afu = ws [(h1 – h4)] – T0 (s1 – s4)]
Rate of exergy loss in the steam generator:
I = afi – afu
The useful mechanical power output:
= Wnet = ws[(h1 – h2) – (h4 – h3)]
Exergy flow rate of the wet steam to the condenser:
afc = ws [(h2 – h3) – T0(s2 – s3)]
Second law efficiency, hII =
Wnet
a f1 - a f2
Exergy flow or Grassman diagram is shown in Fig. 12.28 (c). The energy
disposition diagram (b) shows that the major energy loss (~60%) takes place in the
condenser. This energy rejection, however, occurs at a temperature close to the
ambient temperature, and, therefore, corresponds to a very low exergy value (~4%).
The major exergy destruction due to irreversibilities takes place in the steam
generation. To improve the performance of the steam plant the finite source
temperatures must be closer to the working fluid temperatures to reduce thermal
irreversibility.
12.12
CHARACTERISTICS OF AN IDEAL WORKING FLUID IN
VAPOUR POWER CYCLES
There are certain drawbacks with steam as the working substance in a power cycle.
The maximum temperature that can be used in steam cycles consistent with the best
available material is about 600°C, while the critical temperature of steam is 375°C,
which necessitates large superheating and permits the addition of only an infinitesimal amount of heat at the highest temperature.
High moisture content is involved in going to higher steam pressures in order to
obtain higher mean temperature of heat addition (Tm1). The use of reheat is thus
necessitated. Since reheater tubes are costly, the use of more than two reheats is
hardly recommended. Also, as pressure increases, the metal stresses increase, and
the thicknesses of the walls of boiler drums, tubes, pipe lines, etc. increase not in
proportion to pressure increase, but much faster, because of the prevalence of high
temperature.
Vapour Power Cycles
465
It may be noted that high Tm1 is only desired for high cycle efficiency. High
pressures are only forced by the characteristics (weak) of steam.
If the lower limit is now considered, it is seen that at the heat rejection temperature of 40°C, the saturation pressure of steam is 0.075 bar, which is considerably
lower than atmospheric pressure. The temperature of heat rejection can be still
lowered by using some refrigerant as a coolant in the condenser. The corresponding
vacuum will be still higher, and to maintain such low vacuum in the condenser is a
big problem.
It is the low temperature of heat rejection that is of real interest. The necessity of
a vacuum is a disagreeable characteristic of steam.
The saturated vapour line in the T-s diagram of steam is sufficiently inclined, so
that when steam is expanded to lower pressures (for higher turbine output as well as
cycle efficiency), it involves more moisture content, which is not desired from the
consideration of the erosion of turbine blades in later stages.
The desirable characteristics of the working fluid in a vapour power cycle to
obtain best thermal efficiency are given as follows:
T
(a) The fluid should have a high critical temperature so that the saturation pressure at the maximum permissible temperature (metallurgical limit) is relatively
low. It should have a large enthalpy of evaporation at that pressure.
(b) The saturation pressure at the temperature of heat rejection should be above
atmospheric pressure so as to avoid the necessity of maintaining vacuum in
the condenser.
(c) The specific heat of liquid should be small so that little heat transfer is required to raise the liquid to the boiling point.
(d) The saturated vapour line of the T-s diagram should be steep, very close to the
turbine expansion process so that excessive moisture does not appear during
expansion.
(e) The freezing point of the fluid should be below room temperature, so that it
does not get solidified while flowing through the pipelines.
(f) The fluid should be chemically stable and should not
contaminate the materials of
1
construction at any tempera- 650∞C
ture.
10 bar
(g) The fluid should be nontoxic, 620∞C
noncorrosive, not excessively viscous, and low in
cost.
4
The characteristics of such an
1.8 bar
ideal fluid are approximated in the
2
40∞C
3
T-s diagram as shown in Fig. 12.29.
Some superheat is desired to reduce piping losses and improve
s
turbine efficiency. The thermal efFig. 12.29 T-s diagram of an Ideal Working
ficiency of the cycle is very close
Fluid for a Vapour Power Cycle
to the Carnot efficiency.
466
Engineering Thermodynamics
12.13 BINARY VAPOUR CYCLES
No single fluid can meet all the requirements as mentioned above. Although in the
overall evaluation, water is better than any other working fluid, however, in the high
temperature range, there are a few better fluids, and notable among them are (a)
diphenyl ether, (C6H5)2O, (b) aluminium bromide, Al2Br6, and (c) mercury and other
liquid metals like sodium or potassium. From among these, only mercury has actually been used in practice. Diphenyl ether could be considered, but it has not yet
been used because, like most organic substances, it decomposes gradually at high
temperatures. Aluminium bromide is a possibility and yet to be considered.
When p = 12 bar, the saturation temperature for water, aluminium bromide, and
mercury are 187°C, 482.5°C, and 560°C respectively. Mercury is thus a better fluid in
the high temperature range, because at high temperature, its vaporization pressure
is relatively low. Its critical pressure and temperature are 1080 bar and 1460°C respectively.
But in the low temperature range, mercury is unsuitable, because its saturation
pressure becomes exceedingly low and it would be impractical to maintain such a
high vacuum in the condenser. At 30°C, the saturation pressure of mercury is only
2.7 ¥ 10–4 cm Hg. Its specific volume at such a low pressure is very large, and it
would be difficult to accommodate such a large volume flow.
For this reason, mercury vapour leaving the mercury turbine is condensed at a
higher temperature, and the heat released during the condensation of mercury is
utilized in evaporating water to form steam to operate on a conventional turbine.
Thus in the binary (or two-fluid) cycle, two cycles with different working
fluids are coupled in series, the heat rejected by one being utilized in the other.
The flow diagram of mercury-steam binary cycle and the corresponding T-s diagram are given in Figs. 12.30 and 12.31 respectively. The mercury cycle, a-b-c-d, is a
simple Rankine type of cycle using saturated vapour. Heat is supplied to the mercury in process d-a. The mercury expands in a turbine (process a-b) and is then
condensed in process b-c. The feed pump process, c-d, completes the cycle.
The heat rejected by mercury during condensation is transferred to boil water
and form saturated vapour (process 5-6). The saturated vapour is heated from the
external source (furnace) in the superheater (process 6-1). Superheated steam expands in the turbine (process 1-2) and is then condensed (process 2-3). The
feedwater (condensate) is then pumped (process 3-4), heated till it is saturated
liquid in the economizer (process 4-5) before going to the mercury condenser-steam
boiler, where the latent heat is absorbed. In an actual plant the steam cycle is always
a regenerative cycle, but for the sake of simplicity, this complication has been omitted.
Let m represent the flow rate of mercury in the mercury cycle per kg of steam
circulating in the steam cycle. Then for 1 kg of steam,
Q1 = m(ha – hd) + (h1 – h6) + (h5 – h4)
Q2 = h2 – h3
WT = m(ha – hb) + (h1 – h2)
467
Vapour Power Cycles
Gases
Economiser
1 kg steam
a
m kg Hg
Super
heater
1
Steam
turbine
Drum
Hot
gases
Mercury
turbine
Mercury
condenser
b
Steam boiler
c
6
5
Furnace
2
Steam
condenser
d
Down
comers
Mercury boiler
Header
Fig. 12.30
4
Mercury feed
pump
Water feed
pump
Mercury-steam Plant Flow Diagram
m kg
a
Mercury
cycle
1
d
T
Risers
b
c
5 1 kg
6
Steam
cycle
4
3
2
s
Fig. 12.31
3
Mercury-steam Binary Cycle
WP = m(hd – hc) + (h4 – h3)
hcycle =
Q1 - Q2 WT - WP
=
Q1
Q1
and steam rate =
3600
kg/kW h
WT - WP
468
Engineering Thermodynamics
The energy balance of the mercury condenser-steam boiler gives
m(hb – hc) = h6 – h5
\
m=
h6 - h5
hb - hc
kg Hg/kg H2O
To vaporize one kg of water, seven to eight kg of mercury must condense.
The addition of the mercury cycle to the steam cycle results in a marked
increase in the mean temperature of heat addition to the plant as a whole and consequently the efficiency is increased. The maximum pressure is relatively low.
It may be interesting to note that the concept of the binary vapour cycle evolved
from the need of improving the efficiency of the reciprocating steam engine. When
steam expands up to, say, atmospheric temperature, the resultant volume flow rate
of steam becomes too large for the steam engine cylinder to accommodate. So most
of the early steam engines are found to be non-condensing. The binary cycle with
steam in the high temperature and ammonia or sulphur dioxide in the low temperature range, was first suggested by Professor Josse of Germany in the middle of the
nineteenth century. Steam exhausted from the engine at a relatively higher pressure
and temperature was used to evaporate ammonia or sulphur dioxide which operated
on another cycle. But with the progress in steam turbine design, such a cycle was
found to be of not much utility, since modern turbines can cope efficiently with a
large volume flow of steam.
The mercury-steam cycle has been in actual commercial use for more than three
decades. One such plant is the Schiller Station in the USA. But it has never attained
wide acceptance because there has always been the possibility of improving steam
cycles by increasing pressure and temperature, and by using reheat and regeneration. Over and above, mercury is expensive, limited in supply, and highly toxic.
The mercury-steam cycle represents the two-fluid cycles. The mercury cycle is
called the topping cycle and the steam cycle is called the bottoming cycle. If a
sulphur dioxide cycle is added to it in the low temperature range, so that the heat
released during the condensation of steam is utilized in forming sulphur dioxide
vapour which expands in another turbine, then the mercury-steam-sulphur dioxide
cycle is a three-fluid or tertiary cycle. Similarly, other liquid metals, apart from mercury, like sodium or potassium, may be considered for a working fluid in the topping
cycle. Apart from SO2 other refrigerants (ammonia, freons, etc.) may be considered
as working fluids for the bottoming cycle.
Since the possibilities of improving steam cycles are diminishing, and the
incentives to reduce fuel cost are very much increasing, coupled cycles, like the
mercury-steam cycle, may receive more favourable consideration in the near future.
12.14 THERMODYNAMICS OF COUPLED CYCLES
If two cycles are coupled in series where heat lost by one is absorbed by the other
(Fig. 12.32), as in the mercury-steam binary cycle, let h1 and h2 be the efficiencies of
the topping and bottom cycles respectively, and h be the overall efficiency of the
combined cycle.
469
Vapour Power Cycles
m1
WT2
WT1
T1
B
T2
Q1
m2
Topping
cycle
Q3
Q2
C-B
Fig. 12.32
Bottom
cycle
Two Vapour Cycles Coupled in Series
h1 = 1 –
Q2
Q
and h2 = 1 – 3
Q1
Q2
or
Q2 = Q1(1 – h1) and Q3 = Q2(1 – h2)
Now
h =1–
Q3
Q (1 - h 2 )
=1– 2
Q1
Q1
=1–
Q1 (1 - h1 )(1 - h 2 )
Q1
= 1 – (1 – h1) (1 – h2)
If there are n cycles coupled in series, the overall efficiency would be given by
p
h = 1 – p (1 – hi)
i=1
i.e.
or
h = 1 – (1 – h1) (1 – h2) (1 – h3) … (1 – hn)
1 – h = (1 – h1) (1 – h2) (1 – h3) … (1 – hn)
\ Total loss = Product of losses in all the cycles.
For two cycles coupled in series
h = 1 – (1 – h1)(1 – h2 )
= 1 – (1 – h1 – h2 + h1 h2)
= h1 + h2 – h1 h2
or
h = h1 + h2 – h1 h2
This shows that the overall efficiency of two cycles coupled in series equals the
sum of the individual efficiencies minus their product.
By combining two cycles in series, even if individual efficiencies are low, it is
possible to have a fairly high combined efficiency, which cannot be attained by a
single cycle.
For example, if h1 = 0.50 and h2 = 0.40
h = 0.5 + 0.4 – 0.5 ¥ 0.4 = 0.70
It is almost impossible to achieve such a high efficiency in a single cycle.
470
Engineering Thermodynamics
12.15 PROCESS HEAT AND BY-PRODUCT POWER
There are several industries, such as paper mills, textile mills, chemical factories,
dying plants, rubber manufacturing plants, sugar factories, etc., where saturated
steam at the desired temperature is required for heating, drying, etc. For constant
temperature heating (or drying), steam is a very good medium, since isothermal
condition can be maintained by allowing saturated steam to condense at that temperature and utilizing the latent heat released for heating purposes. Apart from the
process heat, the factory also needs power to drive various machines, for lighting,
and for other purposes.
Formerly it was the practice to generate steam for power purposes at a moderate
pressure and to generate separately saturated steam for process work at a pressure
which gave the desired heating temperature. Having two separate units for process
heat and power is wasteful, for of the total heat supplied to the steam for power
purposes, a greater part will normally be carried away by the cooling water in the
condenser.
By modifying the initial
steam pressure and exhaust
1
WT
pressure, it is possible to
generate the required power
Boiler
and make available for proQ1
cess work the required quantity of exhaust steam at the
Back pressure
desired temperature. In Fig.
turbine
2
12.33, the exhaust steam
QH
from the turbine is utilized
Process
for process heating, the proheater
cess heater replacing the
3
condenser of the ordinary
4
Rankine cycle. The pressure
WP
at exhaust from the turbine is
the saturation pressure corFig. 12.33 Back Pressure Turbine
responding to the temperature desired in the process
heater. Such a turbine is called a back pressure turbine. A plant producing both
power and process heat is sometimes known as a cogeneration plant. When the
process steam is the basic need, and the power is produced incidentally as a byproduct, the cycle is sometimes called a by-product power cycle. Figure 12.34 shows
the T-s plot of such a cycle. If WT is the turbine output in kW, QH the process heat
required in kJ/h, and w is the steam flow rate in kg/h
WT ¥ 3600 = w(h1 – h2 )
and
w(h2 – h3) = QH
QH
\
WT ¥ 3600 =
(h – h )
h2 - h3 1 2
471
Vapour Power Cycles
T
1
Q1
Wp
WT
4
3
2
QH
s
Fig. 12.34 By-product Power Cycle
or
QH =
WT ¥ 3600 ¥ ( h2 - h3 )
kJ/h
h1 - h2
Of the total energy input Q1 (as heat) to the by-product cycle, WT part of it only is
converted into shaft work (or electricity). The remaining energy (Q1 – WT), which
would otherwise have been a waste, as in the Rankine cycle (by the Second Law), is
utilized as process heat.
Fraction of energy (Q1) utilized in the form of work (WT), and process heat (QH) in
a by-product power cycle
=
WT + QH
Q1
Condenser loss, which is the biggest loss in a steam plant, is here zero, and the
fraction of energy utilized is very high.
In many cases the power available from the back pressure turbine through which
the whole of the heating steam flows is appreciably less than that required in the
factory. This may be due to relatively high back pressure, or small heating requirement, or both. Pass-out turbines are employed in these cases, where a certain quantity of steam is continuously extracted for heating purposes at the desired temperature and pressure. (Figs. 12.35 and 12.36).
Q1 = w(h1 – h8)kJ/h
Q2 = (w – w1) (h3 – h4)kJ/h
QH = w1(h2 – h6) kJ/h
WT = w(h1 – h2) + (w – w1) (h2 – h3) kJ/h
WP = (w – w1) (h5 – h4) + w1(h7 – h6) kJ/h
w1h7 + (w – w1)h5 = w ¥ h8
where w is the boiler capacity (kg/h) and w1 is the steam flow rate required (kg/h) at
the desired temperature for process heating.
472
Engineering Thermodynamics
w (kg / h)
1
WT
Boiler
Turbine
Q1
w1(kg / h)
3
(w - w1)
Condenser
2
W
QH
Process
heater
6
w1
Q2
4
wP2
7
(w - w1)
wP1
5
Fig. 12.35 Pass-out Turbine
1
W
T
aw
Q1
7
WT
w1
8
2
6
5
4
(w – w1)
QH
3
Q2
s
Fig. 12.36 T-s Diagram of Power and Process Heat Plant
12.16 EFFICIENCIES IN STEAM POWER PLANT
For the steady flow operation of a turbine, neglecting changes in K.E. and P.E.
(Figs. 12.37 and 12.38).
Maximum or ideal work output per unit mass of steam
(WT)max = (WT)ideal = h1 – h2s
= Reversible and adiabatic enthalpy drop in turbine
473
Vapour Power Cycles
Steam (Ws kg/h)
Blades
WT
Bearings
1
h
1
Brake output
2s 2
Nozzles
2
s
Steam exhaust
Fig. 12.37
Efficiencies in a Steam Turbine
Fig. 12.38
Internal Efficiency of a
Steam Turbine
This work is, however, not obtainable, since no real process is reversible. The expansion process is accompanied by irreversibilities. The actual final state 2 can be
defined, since the temperature, pressure, and quality can be found by actual measurement. The actual path 1–2 is not known and its nature is immaterial, since the
work output is here being expressed in terms of the change of a property, enthalpy.
Accordingly, the work done by the turbine in irreversible adiabatic expansion from
1 to 2 is
(WT)actual = h1 – h2
This work is known as internal work, since only the irreversibilities within the flow
passages of turbine are affecting the state of steam at the turbine exhaust.
\ Internal output = Ideal output – Friction and other losses within the turbine
casing
If ws is the steam flow rate in kg/h
Internal output = ws(h1 – h2) kJ/h
Ideal output = ws (h1 – h2s) kJ/h
The internal efficiency of turbine is defined as
h internal =
h -h
Internal output
= 1 2
Ideal output
h1 - h2 s
Work output available at the shaft is less than the internal output because of the
external losses in the bearings, etc.
\ Brake output or shaft output
= Internal output – External losses
= Ideal output – Internal and External losses
= (kW ¥ 3600 kJ/h)
The brake efficiency of turbine is defined as
hbrake =
Brake output
Ideal output
=
kW ¥ 3600
ws (h1 - h2 s )
474
Engineering Thermodynamics
The mechanical efficiency of turbine is defined as
hmech =
=
Brake output
Internal output
kW ¥ 3600
w s ( h1 - h2 )
\
hbrake = hinternal ¥ hmech
While the internal efficiency takes into consideration the internal losses, and the
mechanical efficiency considers only the external losses, the brake efficiency takes
into account both the internal and external losses (with respect to turbine casing).
The generator (or alternator) efficiency is defined as
Output at generator terminals
Brake output of turbine
The efficiency of the boiler is defined as
hgenerator =
h boiler =
w ( h - h4 )
Energy utilized
= s 1
Energy supplied
w f ¥ C.V.
where wf is the fuel burning rate in the boiler (kg/h) and C.V. is the calorific value of
the fuel (kJ/kg), i.e. the heat energy released by the complete combustion of unit
mass of fuel.
The power plant is an energy converter from fuel to electricity (Fig. 12.39), and
the overall efficiency of the plant is defined as
(Wf ¥ C.V.) kJ / h
Fuel
Fig. 12.39
(kW ¥ 3600) kJ / h
Power
Plant
Electricity
Power Plant–an Energy Converter from Fuel to Electricity
hoverall = hplant =
kW ¥ 3600
w f ¥ C.V.
This may be expressed as follows:
hoverall =
kW ¥ 3600 w s (h1 - h4 ) ws ( h1 - h2 )
=
¥
w f ¥ C.V. ws ( h1 - h4 )
w f ¥ C.V.
¥
or
Brake output
kW ¥ 3600
¥
Brake outpu
w s ( h1 - h2 )
hoverall = hboiler ¥ hcycle ¥ hturbine (mech) ¥ hgenerator
where pump work has been neglected in the expression for cycle efficiency.
475
Vapour Power Cycles
Solved Examples
Example 12.1 Determine the work required to compress steam isentropically
from 1 bar to 10 bar, assuming that at the initial state the steam exists as (a)
saturated liquid and (b) saturated vapour. Neglect changes in kinetic and
potential energies. What conclusion do you derive from this example?
Solution The compression processes are shown in Fig. 12.40.
(a) Steam is a saturated liquid initially, and its specific volume is:
v1 = (vf)1bar = 0.001043 m3/kg
Since liquid is incompressible, v1 remains constant.
2
Wrev = 1 v dp = v1(p1 – p2) = 0.001043 (1 – 10) ¥ 102
z
= – 0.9387 kJ/kg.
2
10 bar
T
(b)
2
1 bar
(a) 1
1
s
Fig. 12.40 Compression of Steam Isentropically
(b) Steam is a saturated vapour initially and remains a vapour during the entire
compression process. Since the specific volume of a gas changes considerably
during a compression process, we need to know how v varies with p to perform the
integration – Ú v dp. This relation is not readily available. But for an isentropic
process, it is easily obtained from the property relation
Tds = dh – vdp = 0
or
v dp = dh
z
2
z
2
Wrev = 1 v dp = – 1 dh = h1 – h2
From steam tables,
h1 = (hg)1bar = 2675.5 kJ/kg
s1 = (sg)1bar = 7.3594 kJ/kg K = s2
For p = 10 bar = 1 MPa and s = 7.3594 kJ/kg K, by interpolation
h2 = 3195.5 kJ/kg
Wrev = 2675.5 – 3195.5 = – 520 kJ/kg
476
Engineering Thermodynamics
It is thus observed that compressing steam in vapour form would require over
500 times more work than compressing it in liquid form for the same pressure rise.
Example 12.2 Steam at 20 bar, 360°C is expanded in a steam turbine to 0.08
bar. It then enters a condenser, where it is condensed to saturated liquid water.
The pump feeds back the water into the boiler. (a) Assuming ideal processes, find
per kg of steam the net work and the cycle efficiency. (b) If the turbine and the
pump have each 80% efficiency, find the percentage reduction in the net work and
cycle efficiency.
Solution The property values at different state points (Fig. 12.41) found from the
steam tables are given below.
h1 = 3159.3 kJ/kg
s1 = 6.9917 kJ/kg K
h3 = h f = 173.88 kJ/kg
s3 = s f = 0.5926 kJ/kg K
p2
p2
= 2403.1 kJ/kg
h fg
p2
vf
p2
s gp2 = 8.2287 kJ/kg K
= 0.001008 m3/kg
Now
s1 = s2s = 6.9917 = s f
\
x 2s =
\
p2
+ x2s s f
gp 2
s fg
p2
= 7.6361 kJ/kg K
= 0.5926 + x2 · 7.6361
6.3991
= 0.838
7.6361
1
360°C
T
p1 = 20 bar
4
4s
p2 = 0.08 bar
3
2s 2
s
Fig. 12.41
\
h2s – hfp2 + x2shfgp2 = 173.88 + 0.838 ¥ 2403.1
= 2187.68 kJ/kg
3
m
kN
(a) WP = h4s – h3 = vfp2 (p1 – p2) = 0.001008
¥ 19.92 ¥ 100 2
kg
m
= 2.008 k/kg
h 4s = 175.89 kJ/kg
WT = h1 – h2s
\
= 3159.3 – 2187.68 = 971.62 kJ/kg
Wnet = WT – WP = 969.61 kJ/kg
Vapour Power Cycles
477
Q1 = h1 – h4s = 3159.3 – 175.89
= 2983.41 kJ/kg
\
hcycle =
(b) If
Wnet
969.61
=
= 0.325, or 32.5%
Q1
2983.41
hP = 80%,
and hT = 80%
2.008
= 2.51 kJ/kg
0.8
WT = 0.8 ¥ 971.62 = 777.3 kJ/kg
\
Wnet = WT – WP = 774.8 kJ/kg
\ % Reduction in work output
WP =
=
969.61 - 774.8
¥ 100 = 20.1%
969.61
h 4s = 173.88 + 2.51 = 176.39 kJ/kg
\
Q1 = 3159.3 – 176.39 = 2982.91 kJ/kg
774.8
= 0.2597, or 25.97%
2982.91
\ % Reduction in cycle efficiency
\
hcycle =
=
0.325 - 0.2597
¥ 100 = 20.1%
0.325
Example 12.3 A cyclic steam power plant is to be designed for a steam temperature at turbine inlet of 360°C and an exhaust pressure of 0.08 bar. After isentropic
expansion of steam in the turbine, the moisture content at the turbine exhaust is
not to exceed 15%. Determine the greatest allowable steam pressure at the turbine
inlet, and calculate the Rankine cycle efficiency for these steam conditions. Estimate also the mean temperature of heat addition.
As state 2s (Fig. 12.42), the quality and pressure are known.
1
T
Solution
4s
0.08 bar
3
2s
x2s = 0.85
s
Fig. 12.42
478
Engineering Thermodynamics
\
s2s = sf + x2s sfg = 0.5926 + 0.85 (8.2287 – 0.5926)
= 7.0833 kJ/kg K
Since
s1 = s2s
\
s1 = 7.0833 kJ/kg K
At state 1, the temperature and entropy are thus known. At 360°C, sg = 5.0526 kJ/kg K,
which is less than s1. So from the table of superheated steam, at t1 = 360°C and s1 =
7.0833 kJ/kg K, the pressure is found to be 16.832 bar (by interpolation).
\ The greatest allowable steam pressure is
p1 = 16.832 bar
h1 = 3165.54 kJ/kg
h 2s = 173.88 + 0.85 ¥ 2403.1 = 2216.52 kJ/kg
h3 = 173.88 kJ/kg
h4s – h3 = 0.001 ¥ (16.83 – 0.08) ¥ 100 = 1.675 kJ/kg
\
h 4s = 175.56 kJ/kg
Q1 = h1 – h4s = 3165.54 – 175.56
= 2990 kJ/kg
WT = h1 – h2s = 3165.54 – 2216.52 = 949 kJ/kg
WP = 1.675 kJ/kg
hcycle =
Wnet 947.32
=
= 0.3168 or 31.68%
Q1
2990
Mean temperature of heat addition
Tm1 =
h1 - h4 s
2990
=
7.0833 - 0.5926
s1 - s4 s
= 460.66 K = 187.51°C
Example 12.4 A steam power station uses the following cycle:
Steam at boiler outlet—150 bar, 550°C
Reheat at 40 bar to 550°C
Condenser at 0.1 bar.
Using the Mollier chart and assuming ideal processes, find the (a) quality at
turbine exhaust, (b) cycle efficiency, and (c) steam rate.
Solution The property values at different states (Fig. 12.43) are read from the
Mollier chart.
h1 = 3465, h2s = 3065, h3 = 3565,
h 4s = 2300 kJ/kg, x4s = 0.88, h5 (steam table) = 191.83 kJ/kg
Quality at turbine exhaust = 0.88
WP = v Dp = 10–3 ¥ 150 ¥ 102 = 15 kJ/kg
h 6s = 206.83 kJ/kg
Q1 = (h1 – h6s) + (h3 – h2s)
479
Vapour Power Cycles
p2 = 4.0 bar
p3 = 0.1 bar
6s
3
550°C 1
1
3
h
T
p1 = 150 bar
r
ba
0
5
r
1
ba
=
p1
40
=
6s
p 2 p3 = 0.1 bar
2s
4s
5
x4s
4s
5
s
550°C
x4s
s
(a)
(b)
Fig. 12.43
= (3465 – 206.83) + (3565 – 3065) = 3758.17 kJ/kg
WT = (h1 – h2s) + (h3 – h4s)
= (3465 – 3065) + (3565 – 2300) = 1665 kJ/kg
Wnet = WT – WP = 1665 – 15 = 1650 kJ/kg
Wnet
1650
=
= 0.4390, or 43.9%
Q1
3758.17
3600
Steam rate =
= 2.18 kg/kW h
1650
Example 12.5 In a single-heater regenerative cycle the steam enters the turbine at 30 bar, 400°C and the exhaust pressure is 0.10 bar. The feed water heater
is a direct-contact type which operates at 5 bar. Find (a) the efficiency and the
steam rate of the cycle and (b) the increase in mean temperature of heat addition,
efficiency and steam rate, as compared to the Rankine cycle (without regeneration). Neglect pump work.
hcycle =
Solution Figure 12.44 gives the flow, T-s, and h-s diagrams. From the steam tables,
the property values at various states have been obtained.
h1 = 3230.9 kJ/kg
s1 = 6.9212 kJ/kg K = s2 = s3
sg at 5 bar = 6.8213 kJ/kg K
Since s2 > sg, the state 2 must lie in the superheated region. From the table for
superheated steam t2 = 172°C, h2 = 2796 kJ/kg
s3 = 6.9212 = sf0.1 bar + x3sfg0.1 bar
= 0.6493 + x37.5009
\
\
6.2719
= 0.836
7.5009
h3 = 191.83 + 0.836 ¥ 2392.8 = 2192.2 kJ/kg
x3 =
480
Engineering Thermodynamics
30 bar 400°C
1 kg
1
Turbine
5 bar
Boiler
0.1 bar
2
3
m kg
Condenser
Heater
7
4
6
Pump
Pump
5
(1 – m) kg
1 kg
(a)
1
400°C
400°C
h
T
1
30 bar
7
6
5
4
1 kg
1 kg
5 bar
7
5
3
2
b
30
kg
m
r
a
kg
5 b (1 – m)
2
(1 – m) kg
m kg
0.1 bar
(1 – m) kg
1 kg
ar
6
4
0.1 b
(1 – m) kg
3
ar
s
s
(b)
(c)
Fig. 12.44
Since pump work is neglected
h4 = 191.83 kJ/kg = h5
h6 = 640.23 kJ/kg = h7
Energy balance for the heater gives
m(h2 – h6) = (1 – m) (h6 – h5)
m(2796 – 640.23) = (1 – m)(640.23 – 191.83)
2155.77 m = 548.4 – 548.4 m
\
\
548.4
= 0.203 kg
2704.17
WT = (h1 – h2) + (1 – m) (h2 – h3)
= (3230.9 – 2796) + 0.797 (2796 – 2192.2)
= 916.13 kJ/kg
m=
Vapour Power Cycles
481
Q1 = h1 – h6 = 3230.9 – 640.23 = 2590.67 kJ/kg
hcycle =
Steam rate =
Tm1 =
Tm1 (without regeneration) =
=
916.13
= 0.3536, or 35.36%
2590.67
3600
= 3.93 kg/kW h
916.13
h1 - h7
2590.67
=
= 511.95 K = 238.8°C
s1 - s7 6.9212 - 1.8607
h1 - h4
s1 - s4
3039.07
6.9212 - 0.6493
= 484.55 K
= 211.4 °C
Increase in Tm1 due to regeneration
= 238.8 – 211.4 = 27.4°C
WT (without regeneration) = h1 – h3
= 3230.9 – 2192.2 = 1038.7 kJ/kg
3600
= 3.46 kg/kW h
1038.7
Increase in steam rate due to regeneration
= 3.93 – 3.46 = 0.476 kg/kW h
Steam rate (without regeneration) =
\
hcycle (without regeneration) =
\
h1 - h3
1038.7
=
3039.07
h1 - h4
= 0.3418 or 34.18%
Increase in cycle efficiency due to regeneration
= 35.36 – 34.18 = 1.18%
Example 12.6 In a steam power plant the condition of steam at inlet to the
steam generator is 20 bar and 300°C and the condenser pressure is 0.1 bar. Two
feedwater heaters operate at optimum temperature. Determine: (a) the quality of
steam at turbine exhaust, (b) net work per kg of steam, (c) cycle efficiency, and (d)
the stream rate. Neglect pump work.
Solution From Fig. 12.19 (a),
h1 = 3023.5 kJ/kg
s1 = 6.7664 kJ/kg K = s2 = s3 = s4
t sat at 20 bar @ 212°C
t sat at 0.1 bar @ 46°C
DtOA = 212 – 46 = 166°C
482
Engineering Thermodynamics
166
= 55°C
3
\ Temperature at which the first heater operates
= 212 – 55 = 157°C @ 150°C (assumed)
Temperature at which the second heater operates = 157 – 55 = 102°C @ 100°C (assumed)
At 0.1 bar,
h f = 191.83, hfg = 2392.8, sf = 0.6493
sg = 8.1502
\
Temperature rise per heater =
At 100°C,
h f = 419.04, hfg = 2257.0, sf = 1.3069, sg = 7.3549
At 150°C,
h f = 632.20, hfg = 2114.3, sf = 1.8418, sg = 6.8379
6.7664 = 1.8418 + x2 ¥ 4.9961
\
\
x2 = 0.986
h2 = 632.2 + 0.986 ¥ 2114.3
= 2716.9 kJ/kg
6.7664 = 1.3069 + x3 ¥ 6.0480
x3 = 0.903
\
or
h3 = 419.04 + 0.903 ¥ 2257.0
h3 = 2457.1 kJ/kg
6.7664 = 0.6493 + x4 ¥ 7.5010
x4 = 0.816
\
h4 = 191.83 +0.816 ¥ 2392.8
= 2144.3 kJ/kg
Since pump work is neglected, h10 = h9, h8 = h7, h6 = h5. By making an energy
balance for the hp heater
(1 – m1)(h9 – h8) = m1(h2 – h9)
Rearranging
m1 =
h9 - h7
213.16
=
= 0.093 kg
h2 - h7 2297.86
By making an energy balance for the lp heater,
(1 – m1 – m2) (h7 – h6) = m2 (h3 – h7)
(1 – 0.093 – m2) (419.04 – 191.83) = m2 (2457.1 – 419.04)
\
m2 = 0.091 kg
\
WT = 1(h1 – h2) + (1 – m1)(h2 – h3)
+ (1 – m1 – m2)(h3 – h4)
483
Vapour Power Cycles
= (3023.5 – 2716.9) + (1 – 0.093)(2716.9 – 2457.1)
+ (1 – 0.093 – 0.091) (2457.1 – 2144.3) = 797.48 kJ/kg
Q1 = h1 – h9 = 3023.5 – 632.2 = 2391.3 kJ/kg
\
hcycle =
Steam rate =
=
WT - WP 797.48
=
= 0.3334 or 33.34%
Q1
2391.3
3600
Wnet
3600
= 4.51 kg/kW h
797.48
Example 12.7 Dry saturated steam at 40 bar expands in a turbine isentropically to the condenser pressure of 0.075 bar. Hot gases available at 2000 K, and
1 atm pressure are used for steam generation and then exhausted at 450 K to the
ambient atmosphere which is at 300 K and 1 atm. The heating rate provided by the
gas stream is 100 MW. Assuming cp of hot gases as 1.1 kJ/kg K, give an exergy
balance of the plant and compare it with the energy balance, and find the second
law efficiency.
Solution
.
Q1 = wg cpg (Ti – Te) = 100 MW
wg = mass flow rate of hot gas
3
=
100 ¥ 10
= 58.7 kg/s
11
. ¥ ( 2000 - 450)
Exergy flow rate of inlet gas
LM T - 1 - 1n T OP
T Q
NT
2000 O
2000
= 58.7 ¥ 1.1 ¥ 300 LM
N 300 - 1 - ln 300 PQ
a f 1 = w g cpg T 0
i
i
0
0
= 73 MW
Exergy flow rate of exhaust gas stream
LM 450 - 1 - ln 450 OP = 1.83 MW
300 Q
N 300
1.83
The exergy loss rate is only about LM
¥ 100 OP or 2.5% of the initial exergy of the
N 73 Q
a f2 = 58.7 ¥ 1.1 ¥ 300
source gas.
The rate of exergy decrease of the gas stream,
a f i = Exergy input rate = 73 – 1.83 = 71.17 @ 71.2 MW
The rate of exergy increase of steam = Exergy utilization rate
a fu = ws [h1 – h4 – T0(s1 – s4)]
484
Engineering Thermodynamics
Now,
h1 = (hg )40 bar = 2801 kJ/kg, h3 = 169 kJ/kg
s3 = s4 = 0.576 kJ/kgK, h4 = 172.8 kJ/kg
s1 = s2 = 6.068 kJ/kgK, h2 = 1890.2 kJ/kg
WT = h1 – h2 = 2801 – 1890.2 = 910.8 kJ/kg
WP = h4 – h3 = 172.8 – 169 = 3.8 kJ/kg
Q1 = h1 – h4 = 2801 – 172.8 = 2628 kJ/kg
Q2 = h2 – h3 = 1890.2 – 169 = 1721 kJ/kg
.
Q1 = 110 MW input
Steam
generator
Flue gas energy
10 MW (9.1%)
1
4
Hot
gas
es
Tgi
40 bar
T Tge
4
Wnet = 34.5 MW
(31.5%)
ST
P
3
1
Condenser
0.075 bar
3
2
2
s
Q2 = 65.5 MW (59.5%)
(a)
(b)
73 MW input
- 39.9 MW
(46.4%)
Steam
generator
P
1.8 MW Flue gas
exergy (2.5%)
ST
34.5 MW
(47.3%)
Condenser
2.8 MW (3.8%)
(c)
Fig. 12.45
(a) T-s Diagram, (b) Energy Distribution Diagram, (c) Exergy
Distribution Diagram
Vapour Power Cycles
485
Wnet = WT – WP = Q1 – Q2 = 907 kJ/kg
.
Q1 = ws ¥ 2628 = 100 ¥ 103 kW
w s = 38 kg/s
a fu = 38 [2801 – 172.8 – 300 (6.068 – 0.576)]
= 37.3 MW
Rate of exergy destruction in the steam generator
= Rate of exergy decrease of gases – Rate of exergy increase of steam.
.
I = afi – afu = 71.2 – 37.3 = 33.9 MW
Rate of useful mechanical
power output
.
Wnet = 38 ¥ 907 = 34.5 MW
Exergy flow rate of wet steam to the condenser
afc = ws [h2 – h3 – T0(s2 – s3)]
= 38 [1890 – 169 – 300 (6.068 – 0.576)] = 2.8 MW
This is the exergy loss to the surroundings.
The energy and exergy balances are shown in Fig. 12.45 (b) and (c). The second law
efficiency is given by
Useful Exergy output 34.5
hII =
=
= 0.473 or 47.3%
73
Exergy input
Example 12.8 In a steam power plant, the condition of steam at turbine inlet is
80 bar, 500°C and the condenser pressure is 0.1 bar. The heat source comprises a
steam of exhaust gases from a gas turbine discharging at 560°C and 1 atm pressure. The minimum temperature allowed for the exhaust gas steam is 450 K. The
mass flow rate of the hot gases is such that the heat input rate to the steam cycle is
100 MW. The ambient condition is given by 300 K and 1 atm. Determine hI, work
ratio and hII of the following cycles: (a) basic Rankine cycle, without superheat,
(b) Rankine cycle with superheat, (c) Rankine cycle with reheat such that steam
expands in the h.p. turbine until it exits as dry saturated vapour, (d) ideal regenerative cycle, with the exit temperature of the exhaust gas steam taken as 320°C,
because the saturation temperature of steam at 80 bar is close to 300°C.
Solution For the first law analysis of each cycle, knowledge of the h values at
each of the states indicated in Fig. 12.46 is required.
(a) Basic Rankine cycle (Fig. 12.46 (a))
By usual procedure with the help of steam tables,
h1 = 2758, h2 = 1817, h3 = 192 and h4 = 200 kJ/kg
WT = h1 – h2 = 941 kJ/kg, Wp = h4 – h3 = 8 kJ/kg
Q1 = h1 – h4 = 2558 kJ/kg, Wnet = 933 kJ/kg
W
933
hI = net =
= 0.365 or 36.5%
Q1
2558
W - WP 933
Work ratio = T
=
= 0.991
941
WT
486
Engineering Thermodynamics
Te =
T 450 K
4
es
Gas
80 bar
Ti = 833 K
Ti = 833 K
Te =
T 450 K
1
4
0.1 bar
3
500∞C
es
Gas
1
80 bar
2
0.1 bar
3
2
s
(a)
s
(b)
1
3
500∞C
80 bar
80 bar
T
2
6
5
0.1 bar
5
1
T
4
2
4
3
0.1 bar
s
(c)
s
(d)
Fig. 12.46
Power output = hI Q1 = 0.365 ¥ 100 = 36.5 MW
LM
N
Exergy input rate = wg cpg ( Ti - T0 ) - T0 ln
Ti
T0
OP
Q
833
100 ¥ 1000
(833 - 300) - 300 ln
833 - 450
300
= 59.3 MW
=
LM
N
OP
Q
36.5
= 0.616 or 61.6%
59.3
(b) Rankine cycle with superheat (Fig. 12.46 (b))
h1 = 3398, h2 = 2130, h3 = 192 and h4 = 200 kJ/kg
WT = 1268 kJ/kg, WP = 8 kJ/kg, Q1 = 3198 kJ/kg
hII =
1260
= 0.394 or 39.4%
3198
1260
Work ratio =
= 0.994
1268
Exergy input rate = 59.3 MW, Wnet = Q1 ¥ hI = 39.4 MW
36.5
hII =
= 0.664 or 66.4%
59.3
hI =
487
Vapour Power Cycles
Improvements in both first law and second law efficiencies are achieved with
superheating. The specific work output is also increased. Therefore, conventional
vapour power plants are almost always operated with some superheat.
(c) Rankine cycle with reheat (Fig. 12.46 (c))
h1 = 3398, h2 = 2761, h3 = 3482, h4 = 2522, h5 = 192 and h6 = 200 kJ/kg
WT1 = 637 kJ/kg, WT2 = 960 kJ/kg
WT = 637 + 960 = 1597 kJ/kg, WP = 8 kJ/kg
Wnet = 1589 kJ/kg, Q1 = 3198 + 721 = 3919 kJ/kg
hI =
Work ratio =
1589
= 0.405 or 40.5%
3919
Wnet 1589
=
= 0.995
1597
WT
Mechanical power output = 100 ¥ 0.405 = 40.5 MW
Exergy input rate = 59.3 MW,
40.5
hII =
= 0.683 or 68.3%
59.3
Compared with basic Rankine cycle, the second law efficiency for the reheat
cycle shows an increase of about 11% [(0.683 – 0.616)/0.616]. Therefore, most of the
large conventional steam power plants in use today operate on the Rankine cycle
with reheat.
(d) Rankine cycle with complete regeneration (Fig. 12.46 (d))
tsat at 0.1 bar = 45.8°C = 318.8 K and
tsat at 80 bar = 295°C = 568 K
hI = hCarnot = 1 –
T3
318.8
=1–
= 0.439 or 43.9%
568.0
T1
Q1 = h1 – h6 = 2758 – 1316 = 1442 kJ/kg
Wnet = Q1 ¥ hI = 1442 ¥ 0.439 = 633 kJ/kg
WP = 8 kJ/kg WT = 641 kJ/kg
633
= 0.988
641
Power output = 0.439 ¥ 100 = 43.9 MW
Work ratio =
Exergy input rate =
100 ¥ 1000
833
(833 - 300) - 300 ln
833 - 593
300
LM
N
OP
Q
= 94.583 MW @ 94.6 MW
43.9
= 0.464 or 46.4%
94.6
The second law efficiency is lower for regeneration because of the more substantial loss of exergy carried by the effluent gas steam at 593 K.
hII =
488
Engineering Thermodynamics
Example 12.9 A certain chemical plant requires heat from process steam at
120ºC at the rate of 5.83 MJ/s and power at the rate of 1000 kW from the generator
terminals. Both the heat and power requirements are met by a back pressure turbine of 80% brake and 85% internal efficiency, which exhausts steam at 120ºC dry
saturated. All the latent heat released during condensation is utilized in the process heater. Find the pressure and temperature of steam at the inlet to the turbine.
Assume 90% efficiency for the generator.
Solution
At 120ºC, hfg = 2202.6 kJ/kg = h2 – h3 (Fig. 12.47)
1
T
Ws
Q1
4
4s
120°C
Ws
2
3
2s
QH
s
Fig. 12.47
A Thermodynamic Syst
QH = ws (h2 – h3) = 5.83 MJ/s
\
ws =
5830
= 2.647 kg/s
2202.6
Wnet =
1000
kJ/s = Brake output
0.9
Now
hbrake =
1000 / 0.9
Brake output
=
= 0.80
Ideal output ws h1 - h2 s
\
h1 – h2s =
1000
= 524.7 kJ/kg
0.9 ¥ 0.8 ¥ 2.647
Again
hinternal =
h1 - h2
= 0.85
h1 - h2s
\
h1 – h2 = 0.85 ¥ 524.7 = 446 kJ/kg
a f
b
g
h2 = hg at 120ºC = 2706.3 kJ/kg
\
h1 = 3152.3 kJ/kg
h2s = h1 – 524.7 = 2627.6 kJ/kg
Vapour Power Cycles
489
= hf + x2s hfg
= 503.71 + x2s ¥ 2202.6
2123.89
= 0.964
2202.6
\
x 2s =
\
s2s = sf + x 2s ◊ sfg = 1.5276 + 0.964 ¥ 5.6020
= 6.928 kJ/kg K
= s1
At state 1,
h1 = 3152.3 kJ/kg
s1 = 6.928 kJ/kg K
From the Mollier chart
p1 = 22.5 bar
t1 = 360ºC
Example 12.10 A certain factory has an average electrical load of 1500 kW
and requires 3.5 MJ/s for heating purposes. It is proposed to install a single-extraction passout steam turbine to operate under the following conditions:
Initial pressure 15 bar.
Initial temperature 300ºC.
Condenser pressure 0.1 bar.
Steam is extracted between the two turbine sections at 3 bar, 0.96 dry, and is
isobarically cooled without subcooling in heaters to supply the heating load. The
internal efficiency of the turbine (in the L.P. Section) is 0.80 and the efficiency of
the boiler is 0.85 when using oil of calorific value 44 MJ/kg.
If 10% of boiler steam is used for auxiliaries calculate the oil consumption per
day. Assume that the condensate from the heaters (at 3 bar) and that from the
condenser (at 0.1 bar) mix freely in a separate vessel (hot well) before being
pumped to the boiler. Neglect extraneous losses.
Solution Let ws be the flow rate of steam (kg/h) entering the turbine, and w the
amount of steam extracted per hour for process heat (Fig. 12.48).
h1 = 3037.3 kJ/kg
h2 = 561.47 + 0.96 ¥ 2163.8
= 2638.7 kJ/kg
s2 = 1.6718 + 0.96 ¥ 5.3201
= 6.7791 kJ/kg K
= s3s
s3s = 6.7791 = 0.6493 + x3s ¥ 7.5009
x 3s =
6.1298
= 0.817
7.5009
490
Engineering Thermodynamics
h3s = 191.83 + 0.817 ¥ 2392.8 = 2146.75 kJ/kg
h2 – h3s = 2638.7 – 2146.75 = 491.95 kJ/kg
1500 kW
Ws kg / h
15 bar 300∞C
1
H.P.
Turbine
Boiler
L.P.
Turbine
3 bar
0.96 dry
2
Ws
0.1 bar
3
Ws - W
Ws - W
Process
heater
G
W
5
Condenser
4
Hot well
(a)
T
1
300∞C
ws kg / h
15 bar
2
w kg / h
7
5
6
4
2s
3 bar
x2 = 0.91
(ws - w) kg / h
3s 3
0.1 bar
s
(b)
Fig. 12.48
h2 – h3 = 0.8 ¥ 491.95 = 393.56 kJ/kg
\
h3 = 2638.7 – 393.56 = 2245.14 kJ/kg
h5 = 561.47 kJ/kg,
h4 = 191.83 kJ/kg
QH = w(h2 – h5) = 2 (2638.7 – 561.47) = 3.5 MJ/s
491
Vapour Power Cycles
\
w=
3500
1.685 kg/s
2077.2
Now
WT = ws(h1 – h2) + (ws – w) (h2 – h3)
= ws(3037.3 – 2638.7) + (ws – 1.685) ¥ 393.56
= ws ¥ 398.6 + ws ¥ 393.56 – 663.15
= 792.16 ws – 663.15
Neglecting pump work
WT = 1500 kJ/s = 792.16 ws – 663.15
\
ws =
2163.15
= 2.73 kg/s = 9828 kg/h
792.16
By making energy balance for the hot well
(ws – w)h4 + wh5 = wsh6
(2.73 – 1.685) 191.83 + 1.685 ¥ 561.47 = 2.73 ¥ h6
200.46 + 946.08 = 2.73 h6
\
h6 = 419.98 kJ/kg @ h7
Steam raising capacity of the boiler = 1.1 ws kg/h, since 10% of boiler steam is
used for auxiliaries.
\
hboiler =
b
11
. ws h1 - h7
g
w f ¥ C. V .
where wf = fuel burning rate (kg/h)
and C.V.= calorific value of fuel = 44 MJ/kg
a
f
11
. ¥ 9828 ¥ 3037.3 - 419.98
w f ¥ 44000
\
0.85 =
or
wf =
11
. ¥ 9828 ¥ 2617.32
= 756.56 kg/h
0.85 ¥ 44000
=
756.56 ¥ 24
= 18.16 tonnes/day
1000
Example 12.11 A steam turbine gets its supply of steam at 70 bar and 450ºC.
After expanding to 25 bar in high pressure stages, it is reheated to 420ºC at the
constant pressure. Next, it is expanded in intermediate pressure stages to an
appropriate minimum pressure such that part of the steam bled at this pressure
heats the feedwater to a temperature of 180ºC. The remaining steam expands from
this pressure to a condenser pressure of 0.07 bar in the low pressure stage. The
isentropic efficiency of the h.p. stage is 78.5%, while that of the intermediate and
l.p. stages is 83% each. From the above data (a) determine the minimum pressure
492
Engineering Thermodynamics
at which bleeding is necessary, and sketch a line diagram of the arrangement of
the plant, (b) Sketch on the T-s diagram all the processes, (c) determine the quantity of steam bled per kg of flow at the turbine inlet, and (d) calculate the cycle
efficiency. Neglect pump work.
Solution Figure 12.49 gives the flow and T-s diagrams of the plant. It would be
assumed that the feedwater heater is an open heater. Feedwater is heated to 180ºC.
So psat at 180ºC @ 10 bar is the pressure at which the heater operates.
1 kg
70 bar, 450∞C
H.P.
Turbine
I.P.
Turbine
L.P.
Turbine
Generator
2
(1 - m) kg 0.07 bar
m kg
10 bar
3
FW heater
(1 - m) kg
450∞C 1
5
T
25 bar,
420∞C
Reheater
4
Condenser
9
(1 - m) kg
7
70 bar
1 kg
25 bar
7
8
6
Pump-2
4
4s
10 bar
(1- m) kg
0.07 bar
5s 5
s
Pump-1
(a)
420∞C
2
2s
m kg
6
8
9
3
(b)
Fig. 12.49
Therefore, the pressure at which bleeding is necessary is 10 bar. From the
Mollier chart
h1 = 3285, h2s = 3010, h3 = 3280, h4s = 3030 kJ/kg
h3 – h4 = 0.83 (h3 – h4s) = 0.83 ¥ 250 = 207.5 kJ/kg
h4 = 3280 – 207.5 = 3072.5 kJ/kg
h5s = 2225 kJ/kg
h4 – h5 = 0.83 (h4 – h5s) = 0.83 ¥ 847.5 = 703.4 kJ/kg
\
h5 = 3072.5 – 703.4 = 2369.1 kJ/kg
h6 = 162.7 kJ/kg
h8 = 762.81 kJ/kg
h1 – h2 = 0.785 (h1 – h2s) = 0.785 ¥ 275 = 215.9 kJ/kg
h2 = 3285 – 215.9 = 3069.1 kJ/kg
Energy balance for the heater gives
m ¥ h4 + (1 – m)h7 = 1 ¥ h8
m ¥ 3072.5 + (1 – m) ¥ 162.7 = 1 ¥ 762.81
493
Vapour Power Cycles
m=
\
hcycle =
600.11
= 0.206 kg/kg steam flow at turbine inlet.
2909.8
bh - h g + bh - h g + a1 - mfbh - h g
bh - h g + bh - h g
1
2
3
1
4
4
8
3
5
2
215.9 + 207.5 + 0.794 ¥ 703.4
=
2522.2 + 210.9
981.9
=
0.3592 or 35.92%
27331
.
Example 12.12 A binary-vapour cycle operates on mercury and steam. Saturated mercury vapour at 4.5 bar is supplied to the mercury turbine, from which it
exhausts at 0.04 bar. The mercury condenser generates saturated steam at 15 bar
which is expanded in a steam turbine to 0.04 bar. (a) Find the overall efficiency of
the cycle. (b) If 50,000 kg/h of steam flows through the steam turbine, what is the
flow through the mercury turbine? (c) Assuming that all processes are reversible,
what is the useful work done in the binary vapour cycle for the specified steam
flow? (d) If the steam leaving the mercury condenser is superheated to a temperature of 300ºC in a superheater located in the mercury boiler, and if the internal
efficiencies of the mercury and steam turbines are 0.85 and 0.87 respectively, calculate the overall efficiency of the cycle. The properties of saturated mercury are
given below
p (bar)
t(ºC)
hf
hg
sf
sg
vf
vg
(kJ/kg)
(kJ/kg K)
(m3/kg)
4.5
450
62.93 355.98 0.1352 0.5397 79.9 ¥ 10–6 0.068
0.04
216.9
29.98 329.85 0.0808 0.6925 76.5 ¥ 10–6 5.178
Solution The cycle is shown in Fig. 12.50.
For the mercury cycle, ha = 355.98
kJ/kg
= sb = sf + xb sfg
= 0.0808 + xb
\
xb =
m kg
450∞C
sa = 0.5397 kJ/kg K
216.9∞C
200.4∞C
Hg
d
c
0.04
bar
b b¢
15 bar 1
(0.6925 – 0.0808)
1 kg
0.4589
= 0.75
0.6117
H2O
hb = 29.98 + 0.75 ¥ 299.87
= 254.88 kJ/kg
(WT)m = ha – hb = 355.98 – 254.88
= 101.1 kJ/kg
(WP )m = hd – hc = 76.5 ¥ 10–6 ¥
4.46 ¥ 100 = 3.41 ¥ 10–2 kJ /kg
4
3
a
1¢
0.04 bar
2¢
2
s
Fig. 12.50
2¢¢
494
Engineering Thermodynamics
\ Wnet = 101.1 kJ/kg
Q1 = ha – hd = 355.98 – 29.98 = 326 kJ/kg
1011
.
W
\ hm = net =
= 0.31 or 31%
326
Q1
For the steam cycle
h1 = 2792.2 kJ/kg
s1 = 6.4448 kJ/kg K = s2 = sf + x2 sfg2
= 0.4226 + x2 (8.4746 – 0.4226)
6.0222
= 0.748
8.0520
h2 = 121.46 + 0.748 ¥ 2432.9 = 1941.27 kJ/kg
x2 =
(WT)St = h1 – h2 = 2792.2 – 1941.27 = 850.93 kJ/kg
(WP)St = h4 – h3 = 0.001 ¥ 14.96 ¥ 100 = 1.496 kJ/kg @ 1.5 kJ/kg
h4 = 121.46 + 1.5 = 122.96 kJ/kg
Q1 = h1 – h4 = 2792.2 – 122.96 = 2669.24 kJ/kg
(Wnet)St = 850.93 –1.5 = 849.43 kJ/kg
\
Wnet
849.43
= 0.318 or 31.8%
=
Q1
2669.24
Overall efficiency of the binary cycle would be
hst =
hoverall = hm + hSt – h m ◊ hSt
= 0.31 + 0.318 – 0.31 ¥ 0.318
= 0.5294 or 52.94%
hoverall can also be determined in, the following way:
By writing the energy balance for a mercury condenser-steam boiler
m(hb – hc) = 1(h1 – h4)
where m is the amount of mercury circulating for 1 kg of steam in the bottom cycle.
h - h4
2669.24
2669.24
\
m= 1
= 11.87 kg
=
=
hb - hc 254.88 - 29.98 224.90
(Q1)total = m(ha – hd ) = 11.87 ¥ 326 = 3869.6 kJ/kg
(WT)total = m(ha – hb) + (h1 – h2)
= 11.87 ¥ 101.1 + 850.93 = 2051 kJ/kg
(WP)total may be neglected
\
hoverall =
WT
2051
=
= 0.53 or 53%
Q1 3869.6
Ans. (a)
If 50,000 kg/h of steam flows through the steam turbine, the flow rate of
mercury wm would be
wm = 50,000 ¥ 11.87 = 59.35 ¥ 10 4 kg/h
Vapour Power Cycles
495
(WT)total = 2051 ¥ 50,000 = 10255 ¥ 104 kJ/h
= 0.2849 ¥ 105 kW = 28.49 MW
Considering the efficiencies of turbines
(WT)m = ha – h¢b = 0.85 ¥ 101.1 = 84.95 kJ/kg
\
\
\
h¢b = 355.98 – 85.94 = 270.04 kJ/kg
m¢ (hb¢ – hc¢) = (h1 – h4)
m¢ = 2669.24 = 11.12 kg
240.06
(Q1)total = m¢(ha – hd) + 1(h1¢ – h1)
= 11.12 ¥ 326 + (3037.3 – 2792.2)
= 3870.22 kJ/kg
s¢1 = 6.9160 = 0.4226 + x¢2 (8.4746 – 0.4226)
x¢2 =
6.4934
= 0.806
8.0520
h¢2 = 121.46 = 0.806 ¥ 2432.9 = 2082.38 kJ/kg
(WT)st = h¢1 – h¢¢2 = 0.87 (3037.3 – 2082.38)
= 830.78 kJ/kg
(WT)total = 11.12 ¥ 85.94 + 830.78 = 1786.43 kJ/kg
Pump work is neglected.
hoverall =
1786.43
= 0.462 or 46.2%
3870.22
Review Questions
12.1 What are the four basic components of a steam power plant?
12.2 What is the reversible cycle that represents the simple steam power plant? Draw
the flow, p–v, T–s and h–s diagrams of this cycle.
12.3 What do you understand by steam rate and heat rate? What are their units?
12.4 Why is Carnot cycle not practicable for a steam power plant?
12.5 What do you understand by the mean temperature of heat addition?
12.6 For a given T2, show how the Rankine cycle efficiency depends on the mean
temperature of heat addition.
12.7 What is metallurgical limit?
12.8 Explain how the quality at turbine exhaust gets restricted.
12.9 How are the maximum temperature and maximum pressure in the Rankine cycle
fixed?
12.10 When is reheating of steam recommended in a steam power plant? How does
the reheat pressure get optimized?
496
Engineering Thermodynamics
12.11 What is the effect of reheat on (a) the specific output, (b) the cycle efficiency,
(c) steam rate, and (d) heat rate, of a steam power plant?
12.12 Give the flow and T–s diagrams of the ideal regenerative cycle. Why is the
efficiency of this cycle equal to Carnot efficiency? Why is this cycle not practicable?
12.13 What is the effect of regeneration on the (a) specific output, (b) mean
temperature of heat addition, (c) cycle efficiency, (d) steam rate and (e) heat
rate of a steam power plant?
12.14 How does the regeneration of steam carnotize the Rankine cycle?
12.15 What are open and closed heaters? Mention their merits and demerits.
12.16 Why is one open feedwater heater used in a steam plant? What is it called?
12.17 How are the number of heaters and the degree of regeneration get optimized?
12.18 Draw the T–s diagram of an ideal working fluid in a vapour power cycle.
12.19 Discuss the desirable characteristics of a working fluid in a vapour power cycle.
12.20 Mention a few working fluids suitable in the high temperature range of a vapour
power cycle.
12.21 What is a binary vapour cycle?
12.22 What are topping and bottoming cycles?
12.23 Show that the overall efficiency of two cycles coupled in series equals the sum
of the individual cycle efficiencies minus their product.
12.24 What is a cogeneration plant? What are the thermodynamic advantages of such
a plant?
12.25 What is a back pressure turbine? What are its applications?
12.26 What is the biggest loss in a steam plant? How can this loss be reduced?
12.27 What is a pass-out turbine? When is it used?
12.28 Define the following: (a) internal work, (b) internal efficiency, (c) brake efficiency (d) mechanical efficiency, and (e) boiler efficiency.
12.29 Express the overall efficiency of a steam plant as the product of boiler, turbine,
generator and cycle efficiencies.
Problems
12.1 For the following steam cycles, find (a) WT in kJ/kg (b) WP in kJ/kg, (c) Q1 in
kJ/kg, (d) cycle efficiency, (e) steam rate in kg/kW h, and (f) moisture at the
end of the turbine process. Show the results in tabular form with your comments.
Boiler Outlet
Condenser
Pressure
Type of Cycle
10 bar, saturated
-do-do-
1 bar
-do-do-
-do10 bar, 300ºC
0.1 bar
-do-
Ideal Rankine Cycle
Neglect WP
Assume 75% pump and turbine
efficiency
Ideal Rankine Cycle
-do(Contd.)
497
Vapour Power Cycles
(Contd.)
Boiler Outlet
Condenser
Pressure
Type of Cycle
150 bar, 600ºC
-do-
-do-do-
-do-
-do-
10 bar, saturated
0.1 bar
-do-
-do-
-do-do-do-
-do-do-do-
-do-
-do-
-doReheat to 600ºC at maximum
intermediate pressure to limit
end moisture to 15%
-do- but with 85% turbine
efficiency
Isentropic pump process ends
on saturated liquid line
-do- but with 80% machine
efficiencies
Ideal regenerative cycle
Single open heater at 110ºC
Two open heaters at 90ºC and
135ºC
-do- but the heaters are
closed heaters
12.2 A geothermal power plant utilizes steam produced by natural means underground. Steam wells are drilled to tap this steam supply which is available at
4.5 bar and 175ºC. The steam leaves the turbine at 100 mm Hg absolute pressure. The turbine isentropic efficiency is 0.75. Calculate the efficiency of the
plant. If the unit produces 12.5 MW, what is the steam flow rate?
12.3 A simple steam power cycle uses solar energy for the heat input. Water in the
cycle enters the pump as a saturated liquid at 40ºC, and is pumped to 2 bar. It
then evaporates in the boiler at this pressure, and enters the turbine as saturated
vapour. At the turbine exhaust the conditions are 40ºC and 10% moisture. The
flow rate is 150 kg/h. Determine (a) the turbine isentropic efficiency, (b) the
net work output (c) the cycle efficiency, and (d) the area of solar collector
needed if the collectors pick up 0.58 kW/m2.
Ans. (c) 2.78%, (d) 18.2 m2
12.4 In a reheat cycle, the initial steam pressure and the maximum temperature are
150 bar and 550ºC respectively. If the condenser pressure is 0.1 bar and the
moisture at the condenser inlet is 5%, and assuming ideal processes, determine
(a) the reheat pressure, (b) the cycle efficiency, and (c) the steam rate.
Ans. 13.5 bar, 43.6%, 2.05 kg/kW h
12.5 In a nuclear power-plant heat is transferred in the reactor to liquid sodium. The
liquid sodium is then pumped to a heat exchanger where heat is transferred to
steam. The steam leaves this heat exchanger as saturated vapour at 55 bar, and
is then superheated in an external gas-fired superheater to 650ºC. The steam
then enters the turbine, which has one extraction point at 4 bar, where steam
flows to an open feedwater heater. The turbine efficiency is 75% and the condenser temperature is 40ºC. Determine the heat transfer in the reactor and in the
superheater to produce a power output of 80 MW.
498
Engineering Thermodynamics
12.6 In a reheat cycle, steam at 500ºC expands in a h.p. turbine till it is saturated
vapour. It is reheated at constant pressure to 400ºC and then expands in a l.p.
turbine to 40ºC. If the maximum moisture content at the turbine exhaust is
limited to 15%, find (a) the reheat pressure, (b) the pressure of steam at the
inlet to the h.p. turbine, (c) the net specific work output, (d) the cycle efficiency, and (e) the steam rate. Assume all ideal processes.
What would have been the quality, the work output, and the cycle efficiency
without the reheating of steam? Assume that the other conditions remain the
same.
12.7 A regenerative cycle operates with steam supplied at 30 bar and 300ºC, and
condenser pressure of 0.08 bar. The extraction points for two heaters (one
closed and one open) are at 3.5 bar and 0.7 bar respectively. Calculate the thermal efficiency of the plant, neglecting pump work.
12.8 The net power output of the turbine in an ideal reheat-regenertive cycle is
100 MW. Steam enters the high-pressure (H.P.) turbine at 90 bar, 550ºC. After
expansion to 7 bar, some of the steam goes to an open heater and the balance is
reheated to 400ºC, after which it expands to 0.07 bar. (a) What is the steam
flow rate to the H.P. turbine? (b) What is the total pump work? (c) Calculate the
cycle efficiency. (d) If there is a 10ºC rise in the temperature of the cooling
water, what is the rate of flow of the cooling water in the condenser? (e) If the
velocity of the steam flowing from the turbine to the condenser is limited to a
maximum of 130 m/s, find the diameter of the connecting pipe.
12.9 A mercury cycle is superposed on the steam cycle operating between the boiler
outlet condition of 40 bar, 400ºC and the condenser temperature of 40ºC. The
heat released by mercury condensing at 0.2 bar is used to impart the latent heat
of vaporization to the water in the steam cycle. Mercury enters the mercury
turbine as saturated vapour at 10 bar. Compute (a) kg of mercury circulated per
kg of water, and (b) the efficiency of the combined cycle.
The property values of saturated mercury are given below
p (bar)
t (ºC)
hf
hg
(kJ/kg)
10
0.2
515.5
277.3
72.23
38.35
363.0
336.55
sf
sg
(kJ/kg K)
0.1478
0.0967
0.5167
0.6385
vf
(m3/kg)
80.9 ¥ 10–6
77.4 ¥ 10–6
vg
0.0333
1.163
12.10 In an electric generating station, using a binary vapour cycle with mercury in
the upper cycle and steam in the lower, the ratio of mercury flow to steam flow
is 10 : 1 on a mass basis. At an evaporation rate of 1,000,000 kg/h for the
mercury, its specific enthalpy rises by 356 kJ/kg in passing through the boiler.
Superheating the steam in the boiler furnace adds 586 kJ to the steam specific
enthalpy. The mercury gives up 251.2 kJ/kg during condensation, and the steam
gives up 2003 kJ/kg in its condenser. The overall boiler efficiency is 85%. The
combined turbine mechanical and generator efficiencies are each 95% for the
mercury and steam units. The steam auxiliaries require 5% of the energy generated by the units. Find the overall efficiency of the plant.
12.11 A sodium-mercury-steam cycle operates between 1000ºC and 40ºC. Sodium
rejects heat at 670ºC to mercury. Mercury boils at 24.6 bar and rejects heat at
0.141 bar. Both the sodium and mercury cycles are saturated. Steam is formed
Vapour Power Cycles
499
at 30 bar and is superheated in the sodium boiler to 350°C. It rejects heat at
0.08 bar. Assume isentropic expansions, no heat losses, and no regeneration
and neglect pumping work. Find (a) the amounts of sodium and mercury used
per kg of steam, (b) the heat added and rejected in the composite cycle per kg
steam, (c) the total work done per kg steam. (d) the efficiency of the composite
cycle, (e) the efficiency of the corresponding Carnot cycle, and (f) the work,
heat added, and efficiency of a supercritical pressure steam (single fluid) cycle
operating at 250 bar and between the same temperature limits.
For mercury, at 24.6 bar, hg = 366.78 kJ/kg
sg = 0.48 kJ/kg K and at 0.141 bar, sf = 0.09
and
sg = 0.64 kJ/kg K, hf = 36.01 and hg = 330.77 kJ/kg
For sodium, at 1000ºC, hg = 4982.53 kJ/kg
At turbine exhaust,h = 3914.85 kJ/kg
At 670ºC, hf = 745.29 kJ/kg
For a supercritical steam cycle, the specific enthalpy and entropy at the turbine
inlet may be computed by extrapolation from the steam tables.
12.12 A textile factory requires 10,000 kg/h of steam for process heating at 3 bar
saturated and 1000 kW of power, for which a back pressure turbine of 70%
internal efficiency is to be used. Find the steam condition required at the inlet
to the turbine.
12.13 A 10,000 kW steam turbine operates with steam at the inlet at 40 bar, 400ºC
and exhausts at 0.1 bar. Ten thousand kg/h of steam at 3 bar are to be extracted
for process work. The turbine has 75% isentropic efficiency throughout. Find
the boiler capacity required.
12.14 A 50 MW steam plant built in 1935 operates with steam at the inlet at 60 bar,
450ºC and exhausts at 0.1 bar, with 80% turbine efficiency. It is proposed to
scrap the old boiler and put in a new boiler and a topping turbine of efficiency
85% operating with inlet steam at 180 bar, 500ºC. The exhaust from the topping turbine at 60 bar is reheated to 450ºC and admitted to the old turbine. The
flow rate is just sufficient to produce the rated output from the old turbine.
Find the improvement in efficiency with the new set up. What is the additional
power developed?
12.15 A steam plant operates with an initial pressure at 20 bar and temperature 400ºC,
and exhausts to a heating system at 2 bar. The condensate from the heating
system is returned to the boiler plant at 65ºC, and the heating system utilizes
for its intended purpose 90% of the energy transferred from the steam it receives. The turbine efficiency is 70%. (a) What fraction of the energy supplied
to the steam plant serves a useful purpose? (b) If two separate steam plants had
been set up to produce the same useful energy, one to generate heating steam at
2 bar, and the other to generate power through a cycle working between 20 bar,
400ºC and 0.07 bar, what fraction of the energy supplied would have served a
useful purpose?
Ans. 91.2%, 64.5%
12.16 In a nuclear power plant saturated steam at 30 bar enters a h.p. turbine and
expands isentropically to a pressure at which its quality is 0.841. At this
pressure the steam is passed through a moisture separator which removes all
500
Engineering Thermodynamics
the liquid. Saturated vapour leaves the separator and is expanded isentropically
to 0.04 bar in l.p. turbine, while the saturated liquid leaving the separator is
returned via a feed pump to the boiler. The condensate leaving the condenser at
0.04 bar is also returned to the boiler via a second feed pump. Calculate the
cycle efficiency and turbine outlet quality taking into account the feed pump
term. Recalculate the same quantities for a cycle with the same boiler and
condenser pressures but without moisture separation.
Ans. 35.5%, 0.824; 35%; 0.716
12.17 The net power output of an ideal regenerative-reheat steam cycle is 80 MW.
Steam enters the h.p. turbine at 80 bar, 500ºC and expands till it becomes
saturated vapour. Some of the steam then goes to an open feedwater heater and
the balance is reheated to 400ºC, after which it expands in the l.p. turbine to
0.07 bar. Compute (a) the reheat pressure, (b) the steam flow rate to the h.p.
turbine, and (c) the cycle efficiency. Neglect pump work.
Ans. 6.5 bar, 58.4 kg/s, 43.7%
12.18 Figure 12.51 shows the arrangement of a steam plant in which steam is also
required for an industrial heating process. The steam leaves boiler B at 30 bar,
320ºC and expands in the H.P. turbine to 2 bar, the efficiency of the H.P. turbine being 75%. At this point one half of the steam passes to the process heater
P and the remainder enters separator S which removes all the moisture. The dry
steam enters the L.P. turbine at 2 bar and expands to the condenser pressure
0.07 bar, the efficiency of the L.P. turbine being 70%. The drainage from the
separator mixes with the condensate from the process heater and the combined
flow enters the hotwell H at 80ºC. Traps are provided at the exist from P and S.
A pump extracts the condensate from condenser C and this enters the hotwell
at 38ºC. Neglecting the feed pump work and radiation loss, estimate the temperature of water leaving the hotwell which is at atmospheric pressure. Also
calculate, as percentage of heat transferred in the boiler, (a) the heat transferred
in the process heater, and (b) the work done in the turbines.
h = 0.70
30 bar, 320∞C
h = 0.75
B
Boiler
H.P
Turbine
L.P
Turbine
2 bar
0.07 bar
P
S
C
Condenser
Traps
80∞C
H
Fig. 12.51
38∞C
Vapour Power Cycles
501
12.19 In a combined power and process plant the boiler generates 21,000 kg/h of
steam at a pressure of 17 bar, and temperature 230ºC. A part of the steam goes
to a process heater which consumes 132.56 kW, the steam leaving the process
heater 0.957 dry at 17 bar being throttled to 3.5 bar. The remaining steam flows
through a H.P. turbine which exhausts at a pressure of 3.5 bar. The exhaust
steam mixes with the process steam before entering the L.P. turbine which develops 1337.5 kW. At the exhaust the pressure is 0.3 bar, and the steam is 0.912
dry. Draw a line diagram of the plant and determine (a) the steam quality at the
exhaust from the H.P. turbine, (b) the power developed by the H.P. turbine, and
(c) the isentropic efficiency of the H.P. turbine.
Ans. (a) 0.96, (b) 1125 kW, (c) 77%
12.20 In a cogeneration plant, the power load is 5.6 MW and the heating load is
1.163 MW. Steam is generated at 40 bar and 500ºC and is expanded isentropically through a turbine to a condenser at 0.06 bar. The heating load is supplied
by extracting steam from the turbine at 2 bar which condensed in the process
heater to saturated liquid at 2 bar and then pumped back to the boiler. Compute
(a) the steam generation capacity of the boiler in tonnes/h, (b) the heat input to
the boiler in MW, and (c) the heat rejected to the condenser in MW.
Ans. (a) 19.07 t/h, (b) 71.57 MW, and (c) 9.607 MW
12.21 Steam is supplied to a pass-out turbine at 35 bar, 350ºC and dry saturated
process steam is required at 3.5 bar. The low pressure stage exhausts at 0.07 bar
and the condition line may be assumed to be straight (the condition line is the
locus passing through the states of steam leaving the various stages of the
turbine). If the power required is 1 MW and the maximum process load is
1.4 kW, estimate the maximum steam flow through the high and low pressure
stages. Assume that the steam just condenses in the process plant.
Ans. 1.543 and 1.182 kg/s
12.22 Geothermal energy from a natural geyser can be obtained as a continuous supply of steam 0.87 dry at 2 bar and at a flow rate of 2700 kg/h. This is utilized in
a mixed-pressure cycle to augment the superheated exhaust from a high pressure turbine of 83% internal efficiency, which is supplied with 5500 kg/h of
steam at 40 bar and 500ºC. The mixing process is adiabatic and the mixture is
expanded to a condenser pressure of 0.10 bar in a low pressure turbine of 78%
internal efficiency. Determine the power output and the thermal efficiency of
the plant.
Ans. 1745 kW, 35%
12.23 In a study for a space project it is thought that the condensation of a working
fluid might be possible at – 40ºC. A binary cycle is proposed, using Refrigerant-12 as the low temperature fluid, and water as the high temperature fluid.
Steam is generated at 80 bar, 500ºC and expands in a turbine of 81% isentropic
efficiency to 0.06 bar, at which pressure it is condensed by the generation of
dry saturated refrigerant vapour at 30ºC from saturated liquid at –40ºC. The
isentropic efficiency of the R-12 turbine is 83%. Determine the mass ratio of
R-12 to water and the efficiency of the cycle. Neglect all losses.
Ans. 10.86; 44.4%
12.24 Steam is generated at 70 bar, 500ºC and expands in a turbine to 30 bar with an
isentropic efficiency of 77%. At this condition it is mixed with twice its mass
of steam at 30 bar, 400ºC. The mixture then expands with an isentropic
502
Engineering Thermodynamics
efficiency of 80% to 0.06 bar. At a point in the expansion where the pressure is
5 bar, steam is bled for feedwater heating in a direct contact heater, which
raises the feedwater to the saturation temperature of the bled steam. Calculate
the mass of steam bled per kg of high pressure steam and the cycle efficiency.
Assume that the L.P. expansion condition line is straight.
Ans. 0.53 kg; 31.9%
12.25 An ideal steam power plant operates between 70 bar, 550ºC and 0.075 bar. It
has seven feedwater heaters. Find the optimum pressure and temperature at
which each of the heaters operate.
12.26 In a reheat cycle steam at 550ºC expands in an h.p. turbine till it is saturated
vapour. It is reheated at constant pressure to 400ºC and then expands in a l.p.
turbine to 40°C. If the moisture content at turbine exhaust is given to be
14.67%, find (a) the reheat pressure, (b) the pressure of steam at inlet to the h.p.
turbine, (c) the net work output per kg, and (d) the cycle efficiency. Assume all
processes to be ideal.
Ans. (a) 20 bar, (b) 200 bar, (c) 1604 kJ/kg, (d) 43.8%
12.27 In a reheat steam cycle, the maximum steam temperature is limited to 500°C.
The condenser pressure is 0.1 bar and the quality at turbine exhaust is 0.8778.
Had there been no reheat, the exhaust quality would have been 0.7592. Assuming ideal processes, determine (a) reheat pressure, (b) the boiler pressure, (c)
the cycle efficiency, and (d) the steam rate.
Ans. (a) 30 bar, (b) 150 bar, (c) 50.51%, (d) 1.9412 kg/kWh
12.28 In a cogeneration plant, steam enters the h.p. stage of a two-stage turbine at
1 MPa, 200ºC and leaves it at 0.3 MPa. At this point some of the steam is bled
off and passed through a heat exchanger which it leaves as saturated liquid at
0.3 MPa. The remaining steam expands in the l.p. stage of the turbine to 40
kPa. The turbine is required to produce a total power of 1 MW and the heat
exchanger to provide a heating rate of 500 kW. Calculate the required mass
flow rate of steam into the h.p. stage of the turbine. Assume (a) steady condition throughout the plant, (b) velocity and gravity terms to be negligible, (c)
both turbine stages are adiabatic with isentropic efficiencies of 0.80.
Ans. 2.457 kg/s
12.29 A steam power plant is designed to operate on the basic Rankine cycle. The
heat input to the boiler is at the rate of 50 MW. The H2O exits the condenser as
saturated liquid and exits the boiler as saturated vapour. The pressure of steam
at boiler exit is 120 bar and the condenser pressure is 0.04 bar. The heat input to
the boiler is provided by a steady stream of hot gases initially at 2200 K and
1 atm. The hot gases exhaust at 600 K and 1 atm to the surroundings which are
at 600 K and 1 atm. Taking the cp of hot gases as 1.1 kJ/kgK, determine (a) the
cycle efficiency, (b) the work output, (c) the power output (in MW), (d) the
required mass flow rate of steam (in kg/h), (e) the specific steam consumption
(in kg/k Wh), (f) the mass flow rate (in kg/h) of the stream of hot gases, (g) the
exergy flux (in MW) of the inlet gases, (h) the exergy loss rate (in MW) with
the exhaust gases, (i) the exergy consumption (in MW) in the steam generation
process, (j) the exergy consumption (in MW) in the condensation process, (i)
the second law efficiency.
Ans. (f) 1.20 ¥ 105 kg/h, (g) 41.5 MW, (h) 3.14 W,
(i) 16.9 MW, (j) 1.40 MW, (i) 48.3%
Vapour Power Cycles
503
12.30 In a cogeneration plant the steam generator provides 106 kg/h of steam at
80 bar, 480ºC, of which 4 ¥ 105 kg/h is extracted between the first and second
turbine stages at 10 bar and diverted to a process heating load. Condensate
returns from the process heating load at 9.5 bar, 120ºC and is mixed with liquid
exiting the lower-pressure pump at 9.5 bar. The entire flow is then pumped to
the steam generator pressure. Saturated liquid at 0.08 bar leaves the condenser.
The turbine stages and the pumps operate with isentropic efficiencies of 86%
and 80%, respectively. Determine (a) the heating load, in kJ/h, (b) the power
developed by the turbine, in kW, (c) the rate of heat transfer in the steam generator, in kJ/h.
Ans. (a) 9.529 ¥ 108 kJ/h, (b) 236,500 kW, (c) 3.032 ¥ 109 kJ/h
12.31 A power plant operates on a regenerative vapour power cycle with one closed
feedwater heater. Steam enters the first turbine stage at 120 bar, 520°C and
expands to 10 bar, where some of the steam is extracted and diverted to a closed
feedwater heater. Condensate exiting the feedwater heater as saturated liquid at
10 bar passes through a trap into the condenser. The feedwater exits the heater
at 120 bar with a temperature of 170°C. The condenser pressure is 0.06 bar. For
isentropic turbine and pump work, determine (a) the thermal efficiency, (b) the
net power developed for a mass flow rate of 106 kg/h into the first-stage turbine.
Ans. (a) 43.4%, (b) 325 MW
12.32 Reconsider the cycle above, and assume that each turbine has an isentropic
efficiency of 82%, the pump efficiency remain 100%.
Ans. (a) 36.8%, (b) 273.3 MW
13
504
Engineering Thermodynamics
Gas Power Cycles
Here gas is the working fluid. It does not undergo any phase change. Engines operating on gas cycles may be either cyclic or non-cyclic. Hot air engines using air as
the working fluid operate on a closed cycle. Internal combustion engines where the
combustion of fuel takes place inside the engine cylinder are non-cyclic heat engines.
13.1
CARNOT CYCLE (1824)
The Carnot cycle (Fig. 13.1 has been discussed in Chapters 6 and 7. It consists of:
Q1
Q1
1
1
c
s=
Wc
T1
T
p
s=
2
2
T=c
Wc
WE
c
WE
4
Q2
3
4
3
T=c
T2
Q2
s
V
(a)
(b)
Fig. 13.1 Carnot Cycle
Two reversible isotherms and two reversible adiabatics. If an ideal gas is
assumed as the working fluid. Then for 1 kg of gas,
v
v
Q1 – 2 = RT1 ln 2 ; W1 – 2 = RT1 ln 2
v1
v1
Q2 – 3 = 0;
W2 – 3 = – cv (T3 – T2)
v
v
Q3 – 4 = RT2 ln 4 ; W3 – 4 = RT2 ln 4
v3
v3
Q4 – 1 = 0;
W4 – 1 = – cv (T1 – T4)
505
Gas Power Cycles
= Â d- W
 dQ
\
cycle
cycle
Now
FG IJ
H K
FT I
v
=G J
v
HT K
v2
T
= 2
v3
T1
and
1
2
4
1
1/( g -1 )
1/( g -1 )
v2
v
v
v
= 1 or 2 = 3
v4
v3
v4
v1
\
Therefore
Q1 = Heat added = RT1 ln
Wnet = Q1 – Q2 = R ln
v2
v1
v2
◊ (T1 – T2)
v1
T - T2
Wnet
= 1
(13.1)
Q1
T1
The large back work (Wc = W4 – 1 ) is a big drawback for the Carnot gas cycle, as in
the case of the Carnot vapour cycle.
\
hcycle =
13.2
STIRLING CYCLE (1827)
The Stirling cycle (Fig. 13.2) consists of:
Two reversible isotherms and two reversible isochores. For 1 kg of ideal gas
v
Q1 – 2 = W1 – 2 = RT1 ln 2
v1
Q2 – 3 = – cv (T2 – T1); W2 – 3 = 0
v
Q3 – 4 = W3 – 4 = – RT2 ln 3
v4
Q4 – 1 = cv (T1 – T2); W4 – 1 = 0
1
c
T=
V=
V=
2
4
c
=
2
c
p
T
T
1
4
c
3
3
V
s
(a)
(b)
Fig. 13.2 Stirling Cycle
T2
T1
506
Engineering Thermodynamics
Due to heat transfers at constant volume processes, the efficiency of the Stirling
cycle is less than that of the Carnot cycle. However, if a regenerative arrangement is
used such that
Q2 – 3 = Q4 – 1 , i.e. the area under 2–3 is equal to the area under 4–1, then the
cycle efficiency becomes
RT1 ln
h=
v
v2
- RT2 ln 3
v1
v4
T - T2
= 1
v2
T1
RT1 ln
v1
(13.2)
So, the regenerative Stirling cycle has the same efficiency as the Carnot cycle.
13.3
ERICSSON CYCLE (1850)
The Ericsson cycle (Fig. 13.3) is made up of:
1
2
1
p1
T1
p
T
4
T
=
P
c
P
=
c
c
c
=
T=
3
2
4
p2
3
V
T2
s
(a)
(b)
Fig. 13.3
Ericsson Cycle
Two reversible isotherms and two reversible isobars.
For 1 kg of ideal gas
p
Q1 – 2 = W1 – 2 = RT1 ln 1
p2
Q2 – 3 = cp (T2 – T1); W2 – 3 = p2 (v3 – v2) = R(T2 – T1 )
Q3 – 4 = W3 – 4 = – RT2 ln
p1
p2
Q4 – 1 = cp (T1 – T4); W4 – 1 = p1 (v1 – v4) = R (T1 – T2)
Since part of the heat is transferred at constant pressure and part at constant temperature, the efficiency of the Ericsson cycle is less than that of the Carnot cycle.
But with ideal regeneration, Q2 – 3 = Q4 – 1 so that all the heat is added from the
external source at T1 and all the heat is rejected to an external sink at T2, the efficiency of the cycle becomes equal to the Carnot cycle efficiency, since
Gas Power Cycles
507
p
RT2 ln 1
Q2
p2
T
h=1–
=1–
=1– 2
(13.3)
p
Q1
T1
RT1 ln 1
p2
The regenerative, Stirling and Ericsson cycles have the same efficiency as the
Carnot cycle, but much less back work. Hot air engines working on these cycles
have been successfully operated. But it is difficult to transfer heat to a gas at high
rates, since the gas film has a very low thermal conductivity. So there has not been
much progress in the development of hot air engines. However, since the cost of
internal combustion engine fuels is getting excessive, these may find a field of use
in the near future.
13.4
AN OVERVIEW OF RECIPROCATING ENGINES
The reciprocating engine, basically a piston-cylinder device, has a wide range of
applications. It is the powerhouse of the vast majority of automobiles, trucks, light
aircraft, ships, electric power generators and so on. The basic components of such
an engine are shown in Fig. 13.4. The piston reciprocates in the cylinder between
two fixed positions called the top
Spark plug or fuel injector
Valve
dead centre (TDC)-the position
of the piston when it forms the
Clearance
volume
smallest volume in the cylinder- Top dead
and the bottom dead centre
center
Bore
(BDC)-the position of the piston
Cylinder wall
when it forms the largest volume
in the cylinder. The distance
Stroke
between the TDC and the BDC is
the largest distance that the pisPiston
ton can travel in one direction,
Bottom
dead center
and it is called the stroke of the
Reciprocating
engine. The diameter of the pismotion
ton is called the bore. The air or
air-fuel mixture is drawn into the
cylinder through the intake
Crank mechanism
valve, and the combustion products are expelled from the cylinRotary motion
der through the exhaust valve.
The minimum volume formed
in the cylinder when the piston is
Fig. 13.4
Nomenclature for reciprocating
at TDC is called the clearance
piston–cylinder engine
volume. The volume displaced
by the piston as it moves between TDC and BDC is called the displacement volume. The ratio of the maximum volume formed in the cylinder to the minimum
(clearance) volume is called the compression ratio rk of the engine
rk = Vmax/V min = VBDC/VTDC
508
Engineering Thermodynamics
Note that rk is a volume ratio and should not be confused with the pressure ratio.
Another term frequently used in regard to reciprocating engines is the mean
effective pressure (mep), as mentioned earlier in Article 3.3 while studying
indicator diagram. It is a fictitions pressure that, if it acted on the piston during the
entire power stroke, would produce the same amount of net work as that produced
during the actual cycle.
Wnet = m.e.p. ¥ piston area ¥ stroke = m.e.p. ¥ displacement volume
or,
m.e.p., pm = (Wnet)/[Vmax – Vmin] (k Pa)
The mean effective pressure can be used as a parameter to compare the performance of reciprocating engines of equal size. The engine with a larger value of
m.e.p. will deliver more net work per cycle and thus will perform better.
Reciprocating engines are classified as spark-ignition (SI) engines or compression-ignition (CI) engines, depending on how the combustion process in the
cylinder is initiated. In SI engines, also called petrol or gasoline engines, the
combustion of the air-fuel mixture is initiated by a spark plug. In CI engines, also
called diesel engines, the air-fuel mixture is self-ignited as a result of compressing
the mixture above its self-ignition temperature.
13.5
AIR STANDARD CYCLES
Internal combustion engines (Fig. 13.4) in which the combustion of fuel occurs in
the engine cylinder itself are non-cyclic heat engines. The temperature due to the
evolution of heat because of the combustion of fuel inside the cylinder is so high
that the cylinder is cooled by water circulation around it to avoid rapid deterioration. The working fluid, the fuel-air mixture, undergoes permanent chemical change
due to combustion, and the products of combustion after doing work are thrown out
of the engine, and a fresh charge is taken. So the working fluid does not undergo a
complete thermodynamic cycle.
To simplify the analysis of I.C. engines, air standard cycles are conceived. In an
air standard cycle, a certain mass of air operates in a complete thermodynamic
cycle, where heat is added and rejected with external heat reservoirs, and all the
processes in the cycle are reversible. Air is assumed to behave as an ideal gas, and
its specific heats are assumed to be constant. These air standard cycles are so conceived that they correspond to the operations of internal combustion engines.
13.6
OTTO CYCLE (1876)
The Otto cycle is the air standard cycle of the SI engine. It is named after NikolausA.
Otto, a German engineer, who first built a successful four-stroke SI engine in 1876
using the cycle proposed by a Frenchman Alphouse Beau de Rochas in 1862. In
most spark-ignition engines, the piston executes four complete strokes within the
cylinder, and the crankshaft completes two revolutions for each thermodynamic
cycle. These engines are called four-stroke internal combustion engines. A schematic of each stroke as well as p-v diagram for an actual four-stroke SI engine is
given in Fig. 13.4 (a).
Initially, the inlet valve opens (I.V.O.) and fresh charge of fuel and air mixture is
drawn into the cylinder. Then both the intake and exhaust valves are closed, and the
piston is at its lowest position (BDC). During the compression stroke, the piston
509
Gas Power Cycles
moves upward, compressing the air-fuel mixture. Shortly before the piston reaches
its highest position (TDC), the spark plug fires and the mixture ignites, increasing
the pressure and temperature of the system. The high pressure gases force the piston down, which in turn forces the crankshaft to rotate, producing a useful work
output during the expansion or power stroke. At the end of this stroke, the piston
is at its lowest position, and the cylinder is filled with the combustion products.
Next the piston moves upward again, purging the exhaust gases through the exhaust
valve (the exhaust stroke), and down a second time, drawing in fresh air-fuel mixture through the intake valve (the intake stroke). It may be noted that the pressure
in the cylinder is slightly above the atmospheric value during the exhaust stroke and
slightly below it during the intake stroke.
2. Compression
1. Induction
I.V.O.
3. Expansion
Both valves closed
Spark
T.D.C. or inner
dead centre
B.D.C. cor outer
dead centre
(a)
(c)
(b)
I.V.O. Inlet valve opens
I.V.C. Inlet valve closes
E.V.O. Exhaust valve opens
E.V.C. Exhaust valve close
4. Exhaust
E.V.O.
I.V.O.
4
Useful work
Spark
Ex
Spark
Compression
pa
Power or
expansion
E.V.C.
ns
io
3
n
C
om
Atmos
press
1
pr
Exhaust
es
s
i on
Exhaust
5
Induction
Pumping work
(d)
(e)
Fig. 13.4(a)
2
.O.
E.V
I.V.C.
Induction
(f)
4 Stroke scroll type
timing diagram
The S.I. Engine: Four-Stroke Cycle
510
Engineering Thermodynamics
In two-stroke engines, all four functions described above are executed in just
two strokes: the power stroke and the compression stroke. In these engines, the
crank case is sealed, and the outward motion of the piston is used to slightly pressurize the air-fuel mixture in the crankcase, as shown in Fig. 13.4 (b). Also, the
intake and exhaust valves are replaced by openings in the lower portion of the
cylinder wall. During the latter part of the power stroke, the piston uncovers first
the exhaust port, allowing the exhaust gases to be partially expelled, and then the
intake port, allowing the fresh air-fuel mixture to rush in and drive most of the
remaining exhaust gases out of the cylinder. This mixture is then compressed as the
piston moves upward during the compression stroke and is subsequently
ignited by a spark plug.
Cylinder: Compression
and ignition
Crankcase: Intake
Fig. 13.4(b)
Cylinder: Intake and
exhaust
Crankcase: Compression
Operating Principle of the two-Stroke Cycle S.I. Engine
The two-stroke engines are generally less efficient than their four-stroke counterparts because of the incomplete expulsion of the exhaust gases and the partial
expulsion of the fresh air-fuel mixture with the exhaust gases. However, they are
relatively simple and inexpensive, and they have high power-to-weight and powerto-volume ratios, which make them suitable for applications requiring small size
and weight such as for motorcycles, scooters, chain saws and lawn mowers.
Advances in several technologies-such as direct fuel injection, stratified charge
combustion, and electronic controls-brought about a renewed interest in two-stroke
engines that can offer high performance and fuel economy while satisfying the future stringent emission requirements.
The thermodynamic analysis of the actual four-stroke or two-stroke cycles
is quite complex. To simplify the analysis; the air standard cycles (Otto cycle) are
conceived. The execution of the Otto cycle in a piston-cylinder device together
with p-v diagram is illustrated in Fig. 13.5 and described in the following
paragraphs.
511
Gas Power Cycles
Fuel-air
mixture
Inlet valve
p
4
Ignition
system
Combustion
products
3
5
Exhaust valve
1
2, 6
I DC
V O DC
(a)
(b)
Fig. 13.5 (a) S.I. Engine (Horizontal) (b) Indicator Diagram
Process 1-2, Intake. The inlet valve is open, the piston moves to the right, admitting
fuel-air mixture into the cylinder at constant pressure.
Process 2-3, Compression. Both the valves are closed, the piston compresses the
combustible mixture to the minimum volume.
Process 3-4, Combustion. The mixture is then ignited by means of a spark, combustion takes place, and there is an increase in temperature and pressure.
Process 4-5, Expansion. The products of combustion do work on the piston which
moves to the right, and the pressure and temperature of the gases decrease.
Process 5-6, Blow-down. The exhaust valve opens, and the pressure drops to the
initial pressure.
Process 6-1, Exhaust. With the exhaust valve open, the piston moves inwards to
expel the combustion products from the cylinder at constant pressure.
The series of processes as described above constitute a mechanical cycle, and
not a thermodynamic cycle. The cycle is completed in four strokes of the piston.
Figure 13.5 (c) shows the air standard cycle (Otto cycle) corresponding to the
above engine. It consists of: Two reversible adiabatics, one reversible isobar, and
one reversible isochore.
Air is compressed in process 1-2 reversibly and adiabatically. Heat is then added
to air reversibly at constant volume in process 2-3. Work is done by air in expanding reversibly and adiabatically in process 3-4. Heat is then rejected by air reversibly at constant volume in process 4-1, and the system (air) comes back to its initial
state. Heat transfer processes have been substituted for the combustion and blowdown processes of the engine. The intake and exhaust processes of the engine cancel each other.
Let m be the fixed mass of air undergoing the cycle of operations as described
above.
Heat supplied Q1 = Q2 – 3 = mcv (T3 – T2)
Heat rejected Q2 = Q4 – 1 = mcv (T4 – T1)
Efficiency h = 1 –
=1–
Q2
mcv ( T4 - T1 )
=1–
mcv ( T3 - T2 )
Q1
T4 - T1
T3 - T2
(13.4)
512
Engineering Thermodynamics
3
3
Q1
WE
pV g = c
p
T
WE
Q1
s=
2
s=
V
c
2
4
c
Wc
V
Q2
Wc
1
V
Process 3-4,
1
=
c
4
Q2
s
Fig. 13.5
Process 1-2,
=
c
FG IJ
H K
Fv I
T
=G J
T
Hv K
v
T2
= 1
v2
T1
3
4
4
3
\
T2
T
= 3
T4
T1
or
T3
T
= 4
T2
T1
(c) Otto Cycle
g -1
g -1
=
FG v IJ
Hv K
g -1
1
2
T3
T
–1= 4 –1
T2
T1
\
FG IJ
H K
T4 - T1
T
v
= 1 = 2
v1
T3 - T2
T2
Fv I
\ From Eq. (13.4), h = 1 – G J
Hv K
g -1
g -1
2
1
or
hotto = 1 –
1
rkg -1
where rk is called the compression ratio and given by
Volume at the beginning of compression
rk =
Volume at the end of compression
=
V1
v
= 1
V2
v2
(13.5)
Gas Power Cycles
513
The efficiency of the air standard Otto cycle is thus a function of the compression
ratio only. The higher the compression ratio, the higher the efficiency. It is independent of the temperature levels at which the cycle operates. The compression ratio
cannot, however, be increased beyond a certain limit, because of a noisy and destructive combustion phenomenon, known as detonation. It also depends upon the
fuel, the engine design, and the operating conditions.
Figure 13.5 (d) shows the effect of
g = 1.67
compression ratio and the specific heat
ratio on the efficiency of Otto cycle. For
g = 1.4
an air standard cycle, air is the working hotto
g = 1.3
fluid, g = 1.4.
We can observe that the thermal efficiency curve is rather steep at low compression ratios but flattens out starting
with a rk of about 8. Therefore, the in8
1.0
rk
crease in thermal efficiency with the compression ratio is not that pronounced at
Fig. 13.5 (d) Effect of rk and g on
high compression ratios. Also, when high
Otto Cycle Efficiency
compression ratios are used, the temperature of the air fuel mixture rises above the self-ignition temperature of the fuel
when the mixture ignites without the spark, causing an early and rapid combustion
of the fuel ahead of the flame front, followed by almost instantaneous burning of
the remaining mixture. This premature ignition of the fuel, called auto-ignition,
produces an audible noise, which is called engine knock or detonation. This autoignition hurts performance and can cause engine damage, thus setting the upper
limit of the compression ratio that can be used in SI engines.
Improvement of the thermal efficiency of gasoline engines by utilizing higher
compression ratios (upto 12) without facing the auto-ignition problem has been
made possible by using gasoline blends that have good antiknock characteristics,
such as gasoline mixed with tetraethyl lead. Tetraethyl lead has been added to gasoline since the 1920s because it is the cheapest method of raising the octane rating,
which is a measure of the engine knock resistance of a fuel. Leaded gasoline, however, has a very undesirable side effect as it forms compounds during the combustion process that are hazardous to health and pollute the environment. Most cars
made since 1975 have been designed to use unleaded gasoline, and the compression ratios had to be lowered to avoid engine knock. As a result the thermal efficiency of car engines has somewhat decreased. However, owing to improvement in
other areas like reduction in overall automobile weight, improved aerodynamic
design etc, today’s cars have better fuel economy.
The second parameter affecting the thermal efficiency of the Otto cycle is the
specific heat ratio g. For a given rk, the use of a monatomic gas such as argon or
helium as the working fluid yields the highest thermal efficiency (Fig. 13.5 d). The
specific heat ratio g and thus the thermal efficiency of the Otto cycle
decreases as the molecules of the working fluid get larger. At room temperature it is
1.4 for air, 1.3 for carbon dioxide and 1.2 for ethane. The working fluid in actual
engines contains larger molecules such as CO2, and g decreases with temperature,
514
Engineering Thermodynamics
because of which the thermal efficiencies of the actual engines are lower than those
of the Otto cycle and vary from 25 to 30 percent.
13.6.1 Work Output
The net work output for an Otto cycle (Fig. 13.5 (c)) can be expressed
p V - p4 V4 p2V2 - p1V1
Wnet = 3 3
g -1
g -1
Now,
v1 V1
=
= rk, or V 1 = V2 rk = V4
v 2 V2
g
FG IJ = r
H K
p 2 p3
V
=
= 1
V2
p1 p 4
g
k
p3 p 4
=
= rp (say)
p2
p1
\
FG
H
I
r
pV Fr r
=
-r + 1J
G
g -1 H r
r
K
Wnet =
IJ
K
p1V1 p3V3 p4V4 p2 V2
+1
g - 1 p1V1 p1V1
p1V1
1 1
g
p k
g
k
p
k
=
Wnet =
13.6.2
k
p1V1
(rp rkg – 1 – rp – rkg – 1 + 1)
g -1
p1V1
(rp – 1) (rkg – 1 – 1)
g -1
(13.6)
Mean Effective Pressure
The mean effective pressure (m.e.p) of the cycle is given by
Net work output
pm =
Swept volume
where swept volume = V1 – V2 = V2 (rk – 1)
p1V1
(rp - 1)(rKg -1 - 1)
g -1
\
pm =
V2 (rK - 1)
=
p1rk ( rp - 1) ( rkg -1 - 1)
(13.7)
(g - 1) ( rk - 1)
Thus, it is seen that the net work output is directly proportional to the pressure
ratio rp. For given values of rk and g, pm increases with rp. For an Otto cycle, an
increase in rk leads to an increase in pm, Wnet and cycle efficiency.
515
Gas Power Cycles
13.7
DIESEL CYCLE (1892)
The limitation on compression ratio in the S.I. engine can be overcome by compressing air alone, instead of the fuel-air mixture, and then injecting the fuel into
the cylinder in spray form when combustion is desired. The CI engine, first proposed by Rudolph Diesel in the 1890s, is very similar to the SI engine, differing
mainly in the method of initiating combustion. In SI engines, a mixture of air and
fuel is compressed during compression stroke, and the compression ratios are limited by the onset of autoignition or engine knock. In CI engines, only air is compressed during the compression stroke. Therefore, diesel engines can operate at
much higher compression ratios, typically between 12 and 24. The spark plug and
carburettor (for mixing fuel and air) are replaced by a fuel injector in diesel engines. The temperature of air after compression must be high enough so that the
fuel sprayed into the hot air burns spontaneously. The rate of burning can, to some
extent, be controlled by the rate of injection of fuel. An engine operating in this way
is called a compression ignition (C.I.) engine. The sequence of processes in the
elementary operation of a C.I. engine, shown in Fig. 13.6, is given below.
4
p
3
Air valve
5
Fuel valve
1
2, 6
Exhaust
valve
V
(a)
Fig. 13.6
(b)
(a) C.I. Engine (b) Indicator Diagram
Process 1-2, intake. The air valve is open. The piston moves out admitting air into
the cylinder at constant pressure.
Process 2-3, Compression. The air is then compressed by the piston to the minimum volume with all the valves closed.
Process 3-4, Fuel injection and combustion. The fuel valve is open, fuel is sprayed
into the hot air, and combustion takes place at constant pressure.
Process 4-5, Expansion. The combustion products expand, doing work on the piston which moves out to the maximum volume.
Process 5-6, Blow-down. The exhaust valve opens, and the pressure drops to the
initial pressure.
Pressure 6-1, Exhaust. With the exhaust valve open, the piston moves towards the
cylinder cover driving away the combustion products from the cylinder at constant
pressure.
The above processes constitute an engine cycle, which is completed in four
strokes of the piston or two revolutions of the crank shaft.
516
Engineering Thermodynamics
Figure 13.7 shows the air standard cycle, called the Diesel cycle, corresponding
to the C.I. engine, as described above. The cycle is composed of:
=
c
Q1
c
3
=
V
3
p
2
WE
T
p
WE
pVg = c
Wc
V
=
c
4
Wc
4
1
Q1
2
Q2
1
Q2
V
s
(a)
(b)
Fig. 13.7
Diesel Cycle
Two reversible adiabatics, one reversible isobar, and one reversible isochore.
Air is compressed reversibly and adiabatically in process 1-2. Heat is then added
to it from an external source reversibly at constant pressure in process 2-3. Air then
expands reversibly and adiabatically in process 3-4. Heat is rejected reversibly at
constant volume in process 4-1, and the cycle repeats itself.
For m kg of air in the cylinder, the efficiency analysis of the cycle can be made as
given below.
Heat supplied,
Q1 = Q2 – 3 = m cp (T3 – T2 )
Heat rejected,
Q2 = Q4 – 1 = m cv (T4 – T1)
Q2
mcv (T4 - T1 )
=1–
Q1
mc p (T3 - T2 )
T4 - T1
\
h=1–
(13.8)
g (T3 - T2 )
The efficiency may be expressed in terms of any two of the following three ratios
V
v
Compression ratio,
rk = 1 = 1
V2
v2
Efficiency
h=1–
Expansion ratio,
re =
V4
v
= 4
V3
v3
Cut-off ratio,
rc =
V3
v
= 3
V2
v2
It is seen that
rk = re ◊ rc
Process 3-4
FG IJ
H K
T4
v
= 3
v4
T3
T4 = T3
g -1
rcg -1
rkg -1
=
1
reg -1
517
Gas Power Cycles
Process 2-3
T2
pv
v
1
= 2 2 = 2 =
T3
p3 v3
rc
v3
\
1
rc
T2 = T3 ◊
Process 1-2
FG IJ
H K
T1
v
= 2
v1
T2
g -1
=
1
rkg -1
T
1
1
= 3 ◊ g -1
rc rk
rkg -1
Substituting the values of T1, T2 and T4 in the expression of efficiency (Eq. 13.8)
\
T1 = T2 ◊
h=1–
r g -1 T 1
T3 ◊ cg -1 - 3 g -1
rk
rc rk
FG
H
g T3 - T3 ◊
\
hDiesel = 1 –
F
GH
1
rc
IJ
K
1 1 rcg - 1
◊
◊
g rkg -1 rc - 1
(13.9)
I
JK
1 rcg - 1
is also greater than unity. Therefore, the efficiency of the
g rc - 1
Diesel cycle is less than that of the Otto cycle for the same compression ratio.
As rc > 1,
13.8
LIMITED PRESSURE CYCLE, MIXED CYCLE OR DUAL
CYCLE
The air standard Diesel cycle does not simulate exactly the pressure-volume variation in an actual compression ignition engine, where the fuel injection is started
before the end of compression stroke. A closer approximation is the limited pressure cycle in which some part of heat is added to air at constant volume, and the
remainder at constant pressure.
Q1
4
4
pVg = c
p
Q1
=c
T
p
3
s
2
=
V
c
Wc
WE
s=
c
=c
Q1
2
5
Q2
1
WE
3
V
=c
5
Wc
1
Q2
V
s
(a)
(b)
Fig. 13.8 Limited Pressure Cycle. Mixed Cycle or Dual Cycle
518
Engineering Thermodynamics
Figure 13.8 shows the p-v and T-s diagrams of the dual cycle. Heat is added
reversibly, partly at constant volume (2-3) and partly at constant pressure (3-4).
Heat supplied
Q1 = mcv (T3 – T2) + mcp (T4 – T3)
Heat rejected
Q2 = mcv (T5 – T1)
Efficiency
h=1–
=1–
Q2
Q1
mcv ( T5 - T1 )
mcv (T3 - T2 ) + mc p (T4 - T3 )
T5 - T1
(13.10)
( T3 - T2 ) + g ( T4 - T3 )
The efficiency of the cycle can be expressed in terms of the following ratios
V
Compression ratio,
rk = 1
V2
V
Expansion ratio,
re = 5
V4
V
Cut-off ratio,
rc = 4
V3
p
Constant volume pressure ratio, rp = 3
p2
It is seen, as before that
rk = rc ◊ re
=1–
or
re =
rk
rc
Process 3-4
V4
T
T p
= 4 3 = 4
p4 T3
V3
T3
T4
T3 =
rc
rc =
Process 2-3
p2 V2
pV
= 3 3
T3
T2
p
T
T2 = T3 2 = 4
p3
rp ◊ rc
Process 1-2
FG IJ
H K
T1
v
= 2
T2
v1
\
T1 =
g -1
=
T4
rp ◊ rc ◊ rkg -1
1
rkg -1
519
Gas Power Cycles
Process 4-5
FG IJ
H K
T5
v
= 4
v5
T4
g -1
=
1
reg -1
rcg -1
rkg -1
Substituting the values of T1 , T2, T3, and T5 in the expression of efficiency
(Eq. 13.8).
\
T5 = T4 ◊
h=1–
r g -1
T4
T4 ◊ cg -1 rk
rp ◊ rc ◊ rkg -1
F T - T I + g FT - T I
GH r r ◊ r JK GH r JK
4
4
4
4
c
\
hDual = 1 –
13.9
p
c
c
rp ◊ rcg - 1
1
(13.11)
rkg -1 rp - 1 + g rp ( rc - 1)
COMPARISON OF OTTO, DIESEL, AND DUAL CYCLES
The three cycles can be compared on the basis of either the same compression ratio
or the same maximum pressure and temperature.
Figure 13.9 shows the comparison of Otto, Diesel, and Dual cycles for the same
compression ratio and heat rejection. Here
1–2–6–5
—Otto cycle
1–2–7–5
—Diesel cycle
1–2–3–4–5 —Dual cycle
3
2
6
v
p=c
4
g
pV = c
7
s=
s=
p
4
=c
c
5
1
V
(a)
c
7
v=
5
1
=
3
2
c
c
T
p
6
s
(b)
Fig. 13.9 Comparison of Otto, Diesel and Dual Cycles for the Same Compression Ratio
For the same Q2, the higher the Q1, the higher is the cycle efficiency. In the
T-s diagram, the area under 2-6 represents Q1 for the Otto cycle, the area under
2-7 represents Q1 for the Diesel cycle, and the area under 2-3-4 represents Q1 for
the Dual cycle. Therefore, for the same rk and Q2
520
Engineering Thermodynamics
hotto > hDual > hDiesel
Figure 13.10 shows a comparison of the three air standard cycles for the same
maximum pressure and temperature (state 4), the heat rejection being also the same.
Here
1–6–4–5
—Otto cycle
1–7–4–5
—Diesel cycle
1–2–3–4–5 —Dual cycle
Q1 is represented by the area under 6-4 for the Otto cycle, by the area under 7-4 for
the Diesel cycle and by the area under 2-3-4 for the Dual cycle in the T-s plot, Q2
being the same.
\
hDiesel > hdual > hotto
This comparison is of greater significance, since the Diesel cycle would definitely
have a higher compression ratio than the Otto cycle.
4
7
4
p
p
T
3
3
V=c
pV g = c
V
7
2
2
6
6
5
1
=
c
V=
=
c
5
c
1
V
s
(a)
(b)
Fig. 13.10 Comparison of Otto, Diesel and Dual Cycles for the Same Maximum
Pressure and Temperature
13.10
LENOIR CYCLE
The Lenoir cycle (Fig. 13.11) consists of the following processes: consant volume
heat addition (1–2); reversible adiabatic expansion (2–3); and constant pressure
heat rejection (3–1). The Lenoir cycle is applicable to pulse jet engines.
2
2
=
c
pVg = C
Q1
V
p
T
3
p=c
Q1
1
3
1
Fig. 13.11
Lenoir Cycle
Q2
V
Q2
s
521
Gas Power Cycles
For 1 kg gas,
Q1 = Cv (T2 – T1 )
Q2 = Cp (T3 – T1 )
hcycle = 1 –
Using
rp =
c p (T3 - T1 )
Q2
g ( T3 - T1 )
=1–
=1–
cv (T2 - T1 )
T2 - T1
Q1
p2 p1v1
pv
p
,
= 2 2 \ T2 = T1 2 = rp ◊ T1
p1 T1
p1
T2
g -1
g
g -1
g
F p I =FG p IJ \T = T F 1 I
T
=G J
GH r JK
T
Hp K Hp K
3
3
2
2
1
3
\
hLenoir = 1 –
g -1
g
2
g -1
g
1 p
p
2
F 1I
=T r G J
Hr K
p
g [T1 rp1/g - T1 ]
T1rp - T1
F r - 1I
=1–g G
H r - 1 JK
1/g
p
(13.12)
p
Thus the cycle efficiency depends on the pressure ratio and the specific heat
ratio.
13.11
ATKINSON CYCLE
Atkinson cycle (Fig. 13.12) is an ideal cycle for an Otto engine exhausting to a
gas turbine. In this cycle the isentropic expansion (3-4) of an Otto cycle is allowed to further expand to the lowest cycle pressure (3-5) so as to increase the
work output.
3
Q1
3
12341 Otto
12351 Atkinson
pVg = C
Q1
p
V
T
4
2
2
=
C
V
4
=
C
p=
Q2
5
1
Q2
1
V
s
Fig. 13.12
For 1 kg gas,
5
C
Atkinson Cycle
Q1 = Cv (T3 – T2)
Q2 = Cp (T5 – T1 )
522
Engineering Thermodynamics
hcycle = 1 –
c p (T5 - T1 )
cv (T3 - T2 )
=1–
Let rk, compression ratio =
v1
v2
re, expansion ratio =
v5
v3
g ( T5 - T1 )
T3 - T2
T2
v
= 1
T1
v2
\
T2 = T1 rk g – 1
(a)
(b)
T3
p
p p
p p
= 3 = 3◊ 5 = 3◊ 1
T2
p2
p5 p2
p5 p2
g
FG IJ = r
H K
Fv I 1
p
=G J =
p
Hv K r
p3
v5
=
p5
v3
g
e
g
\
T3 = T2 ◊ reg ◊
1
2
2
1
g
k
reg
1
1
g–1
g
=
T
r
◊
r
◊
=
T
◊
1 k
e
1
rk
rkg
rkg
F I
GH JK
T5
v
= 3
v5
T3
g -1
=
1
reg -1
1
reg
r
◊
= T1 e
rk reg -1
rk
Substituting T2, T3 and T5 from Eqs. (b), (c) and (d),
r
T1 ◊ e - T1
rk
hAtkinson = 1 – g
reg
- T1 rkg -1
T1
rk
r -r
= 1 – g ge kg
re - rk
\
13.12
T5 = T3 ◊
1
reg -1
(c)
= T1 ◊
(d)
(13.13)
BRAYTON CYCLE
A simple gas turbine power plant is shown in Fig. 13.13. Air is first compressed
adiabatically in process a-b, it then enters the combustion chamber where fuel is
injected and burned essentially at constant pressure in process b-c, and then the
products of combustion expand in the turbine to the ambient pressure in process
c-d and are thrown out to the surroundings. The cycle is open. The state diagram on
the p-v coordinates is shown in Fig. 13.14. Open cycles are used in aircraft, automotive (buses and trucks) and industrial gas turbine installations.
523
Gas Power Cycles
Fuel
Combustion
chamber
c
b
Turbine
Compressor
a
d
Air
Products
of combustion
(exhaust)
Fig. 13.13
A Simple Gas Turbine Plant
p
Air ssion
pre
com
ion
ns
pa
Ex
Combustion
The Brayton cycle is the air standard
c
b
cycle for the gas turbine power plant.
Here air is first compressed reversibly and
adiabatically, heat is added to it reversibly at constant pressure, air expands in
the turbine reversibly and adiabatically,
and heat is then rejected from the air reversibly at constant pressure to bring it to
d
a
the initial state. The Brayton cycle, thereV
fore, consists of:
Two reversible isobars and two reversFig. 13.14 State Diagram of a Gas
Turbine Plant on p-v Plot
ible adiabatics.
The flow, p-v, and T-s diagrams are shown
in Fig. 13.15. For m kg of air
Q1 = heat supplied = m cp (T3 – T2 )
Q2 = heat rejected = m cp (T4 – T1)
\
Cycle efficiency, h = 1 –
=1–
Q2
Q1
T4 - T1
T3 - T2
(13.14)
Now
T2
=
T1
\
FG p IJ
HpK
(g -1)/ g
2
1
T4
T
–1= 3 –1
T2
T1
=
T3
(Since p2 = p3, and p4 = p1)
T4
524
Engineering Thermodynamics
Q1
2
Wc
Heat
exchanger
3
Wnet = WT – Wc
Compressor
Turbine
Wc
Heat
exchanger
1
4
Q2
(a)
p
Q1
2
WT
3
T
pVg = c
p
=
c
2
p=
Wc
4
1
1
c
4
Q2
V
s
(b)
(c)
Brayton Cycle
Fig. 13.15
or
3
T
T4 - T1
= 1 =
T
T3 - T2
2
FG p IJ
Hp K
( g -1)/ g
1
2
=
FG v IJ
Hv K
g -1
2
1
If rk = compression ratio = v1/v2 the efficiency becomes (from Eq. 13.14)
Fv I
h=1–G J
Hv K
g -1
2
1
or
hBrayton = 1 –
1
(13.15)
rkg -1
If rp = pressure ratio = p2/p1 the efficiency may be expressed in the following
form also
FpI
h=1–G J
Hp K
( g -1)/ g
1
2
or
hBrayton = 1 –
1
(rp )( g -1)/g
(13.16)
525
Gas Power Cycles
The efficiency of the Brayton cycle, therefore, depends upon either the compression ratio or the pressure ratio. For the same compression ratio, the Brayton cycle
efficiency is equal to the Otto cycle efficiency.
A closed cycle gas turbine plant (Fig. 13.15) is used in a gas-cooled nuclear
reactor plant, where the source is a high temperature gas-cooled reactor (HTGR)
supplying heat from nuclear fission directly to the working fluid (a gas).
13.12.1
Comparison between Brayton Cycle and Rankine Cycle
Both Rankine cycle and Brayton cycle consist of two reversible isobars and two
reversible adiabatics (Fig. 13.16). While in Rankine cycle, the working fluid undergoes phase change, in Brayton cycle the working fluid always remains in the gaseous phase. Both the pump and the steam turbine in the case of Rankine cycle, and
the compressor and the gas turbine in the case of Brayton cycle operate through the
same pressure difference of p1 and p2. All are steady-flow machines and the work
p2
transfer is given by -
z
v dp. For Brayton cycle, the average specific
p1
volume of air handled by the compressor is less than the same of gas in the gas
turbine (since the gas temperature is much higher), the work done by the gas turbine
is more than the work input to the compressor, so that there is Wnet
available to deliver. In the case of Rankine cycle, the specific volume of water in
the pump is much less than that of the steam expanding in the steam turbine, so
WT >> WP. Therefore, steam power plants are more popular than the gas turbine
plants for electricity generation.
c
p1
p2
b
p1
Rankine
cycle (1-2-3-4)
T
d
1
p1
Brayton
cycle (a-b-c-d)
a
p2
4
p2
3
2
s
Fig. 13.16
Comparison of Rankine Cycle and Brayton Cycle, both Operating
Between the Same Pressures p1 and p2
526
13.12.2
Engineering Thermodynamics
Comparison between Brayton Cycle and Otto Cycle
Brayton and Otto cycles are shown superimposed on the p-v and T-s diagrams in
Fig. 13.17. For the same rk and work capacity, the Brayton cycle (1–2–5–6) handles
a larger range of volume and a smaller range of pressure and temperature than does
the Otto cycle (1-2-3-4).
In the reciprocating engine field, the Brayton cycle is not suitable. A reciprocating engine cannot efficiently handle a large volume flow of low pressure gas, for
which the engine size (p/4 D 2 L) becomes large, and the friction losses also become
more. So the Otto cycle is more suitable in the reciprocating engine field.
3
pVg = c
2
V
T
p
3
5
2
4
V
6
1
=
=c
5
p=
c
p=
c
4
c
1
V
6
s
(a)
(b)
Fig. 13.17
Comparison of Otto and Brayton Cycles
In turbine plants, however, the Brayton cycle is more suitable than the Otto cycle.
An internal combustion engine is exposed to the highest temperature (after the combustion of fuel) only for a short while, and it gets time to become cool in the other
processes of the cycle. On the other hand, a gas turbine plant, a steady flow device,
is always exposed to the highest temperature used. So to protect material, the maximum temperature of gas that can be used in a gas turbine plant cannot be as high as
in an internal combustion engine. Also, in the steady flow machinery, it is more
difficult to carry out heat transfer at constant volume than at constant pressure.
Moreover, a gas turbine can handle a large volume flow of gas quite efficiently. So
we find that the Brayton cycle is the basic air standard cycle for all modern gas
turbine plants.
13.12.3 Effect of Regeneration on Brayton Cycle Efficiency
The efficiency of the Brayton cycle can be increased by utilizing part of the energy
of the exhaust gas from the turbine in heating up the air leaving the compressor in a
heat exchanger called a regenerator, thereby reducing the amount of heat supplied
from an external source and also the amount of heat rejected. Such a cycle is illustrated in Fig. 13.18. The temperature of air leaving the turbine at 5 is higher than
that of air leaving the compressor at 2. In the regenerator, the temperature of air
leaving the compressor is raised by heat transfer from the turbine exhaust. The
527
Gas Power Cycles
maximum temperature to which the cold air at 2 could be heated is the temperature
of the hot air leaving the turbine at 5. This is possible only in an infinite heat exchanger. In the real case, the temperature at 3 is less than that at 5. The ratio of the
actual temperature rise of air to the maximum possible rise is called the effectiveness of the regenerator. For this case,
t3 - t 2
t 5 - t2
e, Effectiveness =
When the regenerator is used in the idealized cycle (Fig. 13.18), the heat supplied
and the heat rejected are each reduced by the same amount, Qx. The mean
temperature of heat addition increases and the mean temperature of heat rejection
decreases because of the use of the regenerator. The efficiency is increased as a
result, but the work output of the cycle remains unchanged. Here,
Wnet = WT – Wc
Wc
C
T
3
2
4
5
1
Q1
Regenerator
6
Q2
(a)
4
T
p=
c
WT
Q1
3
2
p=
Wc
1
c
5
6
Qx
Qx
Q2
s
(b)
Fig. 13.18
Effect of Regeneration on Brayton Cycle
528
Engineering Thermodynamics
Q1 = h4 – h3 = cp (T4 – T3)
Q2 = h6 – h1 = cp (T6 – T1)
WT = h4 – h5 = cp (T4 – T5)
Wc = h2 – h1 = cp (T2 – T1)
h=1–
Q2
T - T1
=1– 6
T4 - T3
Q1
In practice the regenerator is costly, heavy and bulky, and causes pressure losses
which bring about a decrease in cycle efficiency. These factors have to be balanced
against the gain in efficiency to decide whether it is worthwhile to use the regenerator.
Above a certain pressure ratio (p2/p1) the addition of a regenerator causes a loss
in cycle efficiency when compared to the original Brayton cycle. In this situation
the compressor discharge temperature (T2) is higher than the turbine exhaust gas
temperature (T5) (Fig. 13.18). The compressed air will thus be cooled in the regenerator and the exhaust gas will be heated. As a result both the heat supply and heat
rejected are increased. However, the compressor and turbine works remain unchanged. So, the cycle efficiency (Wnet/Q1) decreases.
Let us now derive an expression for the ideal regenerative cycle when the compressed air is heated to the turbine exhaust temperature in the regenerator so that T3
= T5 and T2 = T6 (Fig. 13.17). Therefore,
LM
N
Q2
T -T
T (T2 / T1 ) - 1
=1– 6 1 =1– 1
Q1
T4 - T3
T4 1 - (T5 / T4 )
h=1–
=1–
Since
T2
=
T1
Fp I
GH p JK
2
1
LM
N
T1 T2 1 - (T1 / T2 )
.
T4 T1 1 - (T5 / T4 )
g -1/ g
=
OP
Q
OP
Q
T4
T5
T1 g -1/g
g
(13.17)
T4 p
For a fixed ratio of (T1/T4), the cycle efficiency drops with increasing pressure
ratio.
h=1–
13.12.4 Effect of Irreversibilities in Turbine and Compressor
The Brayton cycle is highly sensitive to the real machine efficiencies of the turbine
and the compressor. Figure 13.19 shows the actual and ideal expansion and compression processes.
Turbine efficiency, hT =
h3 - h4
T - T4
= 3
h3 - h4 s
T3 - T4 s
Compressor efficiency, hC =
h2 s - h1 T2 s - T1
=
h2 - h1
T2 - T1
529
Gas Power Cycles
T
Wnet = WT – WC = (h3 – h4) – (h2 – h1)
Q1 = h3 – h2 and Q2 = h4 – h1
The net output is reduced by the
3
amount [(h4 – h4s) + (h2 – h2s)], and the
c
heat supplied is reduced by the amount
=
p
(h2 – h2s). The efficiency of the cycle will
2
thus be less than that of the ideal cycle.
4
2s
As hT and hC decrease, hcycle also dec 4s
=
p
creases. The cycle efficiency may approach zero even when hT and hC are of
1
the order of 60 to 70%. This is the main
drawback of a gas turbine plant. The mas
chines have to be highly efficient to obFig. 13.19 Effect of Machine Efficiencies
tain reasonable cycle efficiency.
on Brayton Cycle
13.12.5 Effect of Pressure Ratio on the Brayton Cycle
The efficiency of the Brayton cycle is a function of the pressure ratio as given by
the equation
1
h=1–
( g -1)/ g
( rp )
The more the pressure ratio, the more will be the efficiency.
Let T1 = the lowest temperature of the cycle, which is the temperature of the
surroundings (Tmin), and T3 = the maximum or the highest temperature of the cycle
limited by the characteristics of the material available for burner and turbine construction (Tmax).
Since the turbine, a steady-flow machine, is always exposed to the highest temperature gas, the maximum temperature of gas at the inlet to the turbine is limited to
about 800°C by using a high air-fuel ratio. With turbine blade cooling, however, the
maximum gas inlet temperature can be 1100°C or even higher.
Figure 13.20 shows the Brayton cycles operating between the same Tmax and
Tmin at various pressure ratios. As the pressure ratio changes, the cycle shape also
changes. For the cycle 1-2-3-4 of low pressure ratio rp , since the average temperature of heat addition
3¢
3¢¢
3
Tmax
T
2¢
4
2¢¢
4¢¢
2
4¢
Tmin
1
s
Fig. 13.20 Effect of Pressure Ratio on Brayton Cycle
530
Engineering Thermodynamics
h3 - h2
s3 - s 2
is only a little greater than the average temperature of heat rejection
h -h
Tm2 = 4 1
s4 - s1
the efficiency will be low. At the lower limit of unity pressure ratio, both work
output and efficiency will be zero.
As the pressure ratio is increased, the efficiency steadily increases, because Tm1
increases and Tm2 decreases. The mean
temperature of heat addition Tm1
hcarnot
approaches Tmax and the mean temperature of heat rejection Tm2 approaches Tmin, with the increase in rp .
In the limit when the compression process ends at Tmax, the Carnot efficiency
is reached, rp has the maximum value
(rpmax), but the work capacity again
become zero.
Figure 13.21 shows how the cycle
efficiency varies with the pressure ra0
1
rp
(rp)max
tio, with rp varying between two limiting values of 1 and (rp)max when the Fig. 13.21 Effect of Pressure Ratio on
Carnot efficiency is reached. When rp
Brayton Cycle Efficiency
= (rp)max, the cycle efficiency is given
by
h
Tm1 =
1
h=1–
=1–
\
((rp)max)(g – 1)/g =
\
(rp)max =
(g -1)/ g
((rp ) max
= hCarnot
Tmin
Tmax
Tmax
Tmin
FT I
GH T JK
max
g /( g -1)
(13.18)
min
From Fig. 13.20 it is seen that the work capacity of the cycle, operating between
Tmax and Tmin, is zero when rp = 1 passes through a maximum, and then again becomes zero when the Carnot efficiency is reached. There is an optimum value of
pressure ratio (rp)opt at which work capacity becomes a maximum, as shown in
Fig. 13.22.
For 1 kg,
Wnet = cp [(T3 – T4) – (T2 – T1)]
where
T3 = Tmax and T1 = Tmin
531
Gas Power Cycles
Wnet
(Wnet)max
0
1
(rp)opt
(rp)max
rp
Fig. 13.22
Effect of Pressure Ratio on Net Output
Now
T3
= (rp)(g – 1)/g
T4
\
T4 = T3 ◊ rp– (–g –1)/g
Similarly
T2 = T1 ◊ rp(g – 1)/g
Substituting in the expression for Wnet
Wnet = cp [T3 – T3 ◊ (rp)–(g – 1)/g – T1 rp(g – 1)/g + T1]
To find (rp)opt
LM FG
N H
F g - 1IJ r
T G
H g K
IJ
K
FG
H
(13.19)
OP
Q
IJ
K
g - 1 ( -1+ (1/g ) -1)
g - 1 (1- (1/g ) -1)
dWnet
- T1
rp
rp
= cp - T3 =0
g
g
drp
\
\
(1/g )–2
3
p
= T1
rp–(1/g) – (1/g) + 2 =
FG g - 1IJ ◊ r
H g K
–1/g
p
T3
T1
g / 2( g -1)
FT I
(r ) = G J
HT K
FT I
(r ) = G
H T JK
3
p opt
1
\
g / 2 (g -1)
max
(13.20)
p opt
min
From Eqs. (13.18) and (13.20)
(rp)opt =
(13.21)
( rp ) max
Substituting the values of (rp )opt in Eq. (13.19)
LM
MN
F T I g . g -1
GH T JK 2(g - 1) g
F T I g . g - 1 + T OP
-T G J
H T K 2(g - 1) g PQ
Wnet = (Wnet)max = cp T - T
3
1
3
3
3
1
1
1
532
Engineering Thermodynamics
= cp [T3 – 2 T1T3 + T1]
or
(Wnet)max = cp ( Tmax - Tmin )
hcycle = 1 –
1
g -1/ g
rp
=1–
2
Tmin
Tmax
(13.22)
(13.23)
Considering the cycles 1-2¢-3¢-4¢ and 1-2≤-3≤-4≤ (Fig. 13.21), it is obvious that to
obtain a reasonable work capacity, a certain reduction in efficiency must be accepted.
13.12.6 Effect of Intercooling and Reheating on Brayton Cycle
The efficiency of the Brayton cycle may often be increased by the use of staged
compression with intercooling, or by using staged heat supply, called reheat.
Let the compression process be divided into two stages. Air, after being compressed in the first stage, is cooled to the initial temperature in a heat exchanger,
called an intercooler, and then compressed further in the second stage (Fig. 13.23).
Fig. 13.23
Effect of Intercooling on Brayton Cycle
1-2¢-5-6 is the ideal cycle without intercooling, having a single-stage compression,
1-2-3-4-5-6 is the cycle with intercooling, having a two-stage compression. The cycle
2-3-4-2¢ is thus added to the basic cycle 1-2¢-5-6. There is more work capacity, since
the included area is more. There is more heat supply also. For the cycle 4-2¢-2-3, Tm1
533
Gas Power Cycles
is lower and Tm2 higher (lower rp) than those of the basic cycle 1-2¢-5-6. So the efficiency of the cycle reduces by staging the compression and intercooling. But if a
regenerator is used, the low temperature heat addition (4-2¢) may be obtained by
recovering energy from the exhaust gases from the turbine. So there may be a net gain
in efficiency when intercooling is adopted in conjunction with a regenerator.
Similarly, let the total heat supply be given in two stages and the expansion
process be divided in stages in two turbines (T1 and T2) with intermediate reheat, as
shown in Fig. 13.24. 1-2-3-4¢ is the cycle with a single-stage heat supply having no
reheat, with total expansion in one turbine only. 1-2-3-4-5-6 is the cycle with a
single-stage reheat, having the expansion divided into two stages. With the basic
cycle, the cycle 4-5-6-4¢ is added because of reheat. The work capacity increases,
but the heat supply also increases.
In the cycle 4-5-6-4¢, rp is lower than in the basic cycle 1-2-3-4¢, so its efficiency
is lower. Therefore, the efficiency of the cycle decreases with the use of reheat. But
T6 is greater than T 4¢ . Therefore, if regeneration is employed, there is more energy
that can be recovered from the turbine exhaust gases. So when regeneration is employed in conjunction with reheat, there may be a net gain in cycle efficiency.
Reheater
Q1
5
4
3
2
C
Q1¢
T2¢
T1
Wc
1
Q2
6
(a)
2
=
c
p
g
pV = c
3
p
T
p
2
5
3
=
p
c
=
c
4
6
5
4
4¢
1
1
4¢
V
s
(b)
(c)
Fig. 13.24 Effect of Reheat on Brayton Cycle
6
534
Engineering Thermodynamics
Isotherm
P
If in one cycle, several stages of
intercooling and several stages of
reheat are employed, a cycle as
shown in Fig. 13.25 is obtained.
When the number of such stages is
large the cycle reduces to the
Ericsson cycle with two reversible
isobars and two reversible isotherms. With ideal regeneration the
cycle efficiency becomes equal to
the Carnot efficiency.
Isentropes
Isotherm
Isentropes
V
Fig. 13.25 (a)
=
p
c
=
p
=
p
c
c
=
p
c
c
=
p
T
Isotherm
Isotherm
s
Fig. 13.25(b) Brayton Cycle with Many Stages of Intercooling and Reheating
Approximates to Ericsson Cycle
13.12.7 Ideal Regenerative Cycle with Intercooling and Reheat
Let us consider an ideal regenerative gas turbine cycle with two-stage compression
and a single reheat. It assumes that both intercooling and reheating take place at the
root mean square of the high and low pressure in the cycle, so that p3 = p2 = p7 = p8
=
p1 p4 =
p6 p9 (Fig. 13.26). Also, the temperature after intercooling is equal to
the compressor inlet temperature (T1 = T3) and the temperature after reheat is equal
to the temperature entering the turbine initially (T6 = T8).
Here,
Q1 = cp (T6 – T5) + cp (T8 – T7)
Since
p6
p
= 8 and T6 = T8, it follows that T5 = T7 = T9 .
p7
p9
\
Again,
Q1 = 2 cp (T6 – T7)
Q2 = cp (T10 – T1) + cp (T2 – T3)
but
p2
p
= 4
p1
p3
so that
\
T2 = T4 = T10
Q2 = 2 cp (T2 – T1)
and
T3 = T1
535
Gas Power Cycles
Wnet = Q1 – Q2 = 2 cp [(T6 – T7) – (T2 – T1)]
\
hcycle =
Wnet
T -T
=1– 2 1
Q1
T6 - T7
=1–
LM
N
T1 ( T2 / T1 ) - 1
T6 1 - ( T7 / T6 )
OP
Q
Cooling Water
Q1
Intercooler
3
7
2
C1
Hot
Gas
C2
Reheater
T1
8
T2
6
1
4
9
Q1
5
10
Regenerator
Q2
(a)
8
T
6
5
7
4
2
3
9
10
1
s
(b)
Fig. 13.26
Ideal Regenerative Gas Turbine Cycle with Two-stage
Compression and Reheat
536
Engineering Thermodynamics
=1–
LM
N
T1 T2 1 - ( T1 / T2 )
¥
T6 T1 1 - ( T7 / T6 )
F I
GH JK
T
p
=1– 1 ¥ 2
T6
p1
g -1/ g
p1 p4 ,
But
p2 =
\
hcycle = 1 –
13.12.8
OP
Q
T1
T6
Fp I
GH p JK
(g -1)/ 2g
4
(13.24)
1
Free-shaft Turbine
So far only a single shaft has been shown in flow diagrams, on which were mounted
all the compressors and the turbines. Sometimes, for operating convenience and
part-load efficiency, one turbine is used for driving the compressor only on one
shaft, and a separate turbine is used on another shaft, known as free-shaft, for supplying the load, as shown in Fig. 13.27.
Q1
C
T
W C = WT
Free shaft
turbine
(WT)FST
Q2
Load
Fig. 13.27
Free Shaft Turbine
13.13 AIRCRAFT PROPULSION
Gas turbines are particularly suited for aircraft propulsion because they are light
and compact and have a high power-to-weight ratio. Aircraft gas turbines operate
on an open cycle called “jet propulsion cycle”, as explained below. It can be of
turbojet, turbofan or turborop type. In a turbojet engine (Fig. 13.28), high velocity
air first flows through a diffuser where it is decelerated increasing its pressure.
Air is then compressed in the compressor. It is mixed with fuel in the combustion
chamber, where the mixture is burned at constant pressure. The high pressure,
high temperature combustion gases partially expand in the turbine, producing
enough power to drive the compressor and any auxiliary equipment. Finally, the
gases expand in the nozzle to the ambient pressure and leave the aircraft at high
velocity. In the ideal case, the turbine work is assumed to be equal to the compressor work. The processes in the diffuser, the compressor, the turbine, and the
537
Gas Power Cycles
nozzle are assumed to be reversible and adiabatic. Another view of a turbojet engine is given in Fig. 13.28 (a).
The thrust developed in a turbojet engine is the unbalanced force caused by the
difference in the momentum of the air entering the engine and the exhaust gases
leaving the engine, so that,
∑
∑
∑
F = ( m V ) exit - ( m V ) inlet = m( Vexit - Vinlet )
T
(13.25)
4
st.
n
Q1
p
=
co
2
5
3
3
1
6
4
5
6
Q2
2
1
p = const.
s
Diffuser Compressor Burner sectionTurbine Nozzle
Basic Components of a Turbojet Engine and the T-s Diagram of
an Ideal Turbojet Cycle
Fig. 13.28
Combustors
Compressor
3
T
Turbine
4
Product
gases
out
Air
in
2
5
1
a
1
a
Diffuser
Fig. 13.28(a)
2
3
Gas generator
4
5
Nozzle
s
Turbojet Engine Schematic and Accompanying Ideal T-s Diagram
The pressures at inlet and exit of the engine are the ambient pressure. For an
aircraft cruising in still air, V inlet is the aircraft velocity. The mass flow rates of
the gases at the engine exit and the inlet are different, the difference being equal to
the combustion rate of the fuel. But the air-fuel ratio used in jet propulsion
.
engines is usually very high, making the difference very small. Thus m in
Eq. (13.25) is taken as the mass flow rate of air through the engine. For an aircraft
cruising at a steady speed, the thrust is used to overcome the fluid drag, and the net
force acting on the body of the aircraft is zero. Commercial airplanes save fuel by
flying at higher altitudes during long trips since the air at higher altitudes is of less
density and exerts a smaller drag force on the aircraft.
538
Engineering Thermodynamics
The power developed from the thrust of the engine is called the propulsive power,
Wp, given by:
.
.
W p = FVaircraft = m ( Vexit - Vinlet )Vaircraft
(13.26)
The propulsive efficiency, h p, is defined by:
.
Propulsive power
Wp
hp =
= .
Energy input rate
Q
(13.27)
in
It is a measure of how efficiently the energy released during combustion is converted to propulsive power.
Space and weight limitations prohibit the use of regenerators and intercoolers on
aircraft engines. The counterpart of reheating is afterburning. The air-fuel
ratio in a jet engine is so high that the turbine exhaust gases are sufficiently rich in
oxygen to support the combustion of more fuel in an afterburner (Fig. 13.29). Such
burning of fuel raises the temperature of the gas before it expands in the nozzle,
increasing the K.E. change in the nozzle and consequently increasing the thrust. In
the air-standard case, the combustion is replaced by constant pressure heat addition.
Inlet
Compressor Burner Turbine
Afterburner
Fuel injectors
Fig. 13.29
Exhaust
nozzle
Flame holder
Turbojet Engine with Afterburner
The most widely used engine in aircraft propulsion is the turbofan engine
wherein a large fan driven by the turbine forces a considerable amount of air through
a duct (cowl) surrounding the engine (Figs. 13.30 and 13.31). The fan
Low-pressure
compressor
Fan
Low-pressure
turbine
Duct Burners
Fan exhaust
Turbine
exhaust
Fan
High-pressure
compressor
High-pressure
turbine
Fig. 13.30 Turbofan Engine
exhaust leaves the duct at a higher velocity, enhancing the total thrust of the
engine significantly. Some of the air entering the engine flows through the compressor, combustion chamber and turbine, and the rest passes through the fan into a
539
Gas Power Cycles
duct and is either mixed with the exhaust gases or is discharged separately. It improves the engine performance over a broad operating range. The ratio of the mass
flow rates of the two streams is called the bypass ratio.
Combustion
chamber
Rotor 2
Rotor 1
(a)
Combustion
chamber
Fan
Rotor 2
Rotor 1
(b)
Fig. 13.31 Turbofan or Bypass Jet Engines
.
.
m total - m turbine
Bypass ratio =
.
m turbine
The bypass ratio can be varied in flight by various means. Turbofan engines
deserve most of the credit for the success of jumbo jets, which weigh almost 400,000
kg and are capable of carrying 400 passengers for up to 10,000 km at speeds over
950 km/h with less fuel per passenger mile.
Increasing the bypass ratio of a turbofan engine increases thrust. If the cowl is
removed from the fan the result is a turboprop engine (Fig. 13.32). Turbofan and
Propeller
Compressor
Burner
Turbine
Gear reduction
Fig. 13.32
Turboprop Engine
turboprop engines differ mainly in their bypass ratio: 5 or 6 for turbofans and as
high as 100 for turboprop. In general, propellers are more efficient than jet
engines, but they are limited to low-speed and low-altitude operation since their
efficiency decreases at high speeds and altitudes.
540
Engineering Thermodynamics
A particularly simple type of engine known as a ramjet is shown in Fig. 13.33.
This engine requires neither a compressor nor a turbine. A sufficient pressure rise is
obtained by decelerating the high speed incoming air in the diffuser (ram effect) on
being rammed against a barrier. For the ramjet to operate, the aircraft must already
be in flight at a sufficiently great speed. The combustion products exiting the combustor are expanded through a nozzle to produce the thrust.
Fuel nozzles or spray bars
Jet nozzles
Air inlet
Flameholders
Fig. 13.33 Ramjet Engine
In each of the engines mentioned so far, combustion of the fuel is supported by
air brought into the engines from the atmosphere. For very high altitude flights and
space travel, where this is no longer possible, rockets may be employed. In a rocket,
both fuel and an oxidizer (such as liquid oxygen) are carried on board of the craft.
High pressure combustion gases are expanded in a nozzle. The gases leave the
rocket at very high velocities, producing the thrust to propel the rocket.
13.14
BRAYTON-RANKINE COMBINED CYCLE
Both Rankine cycle and Brayton cycle consist of two reversible isobars and two
reversible adiabatics. While the former is a phase change cycle, in the latter the
working fluid does not undergo any phase change.
A gas turbine power plant operating on Brayton cycle has certain disadvantages
like large compressor work, large exhaust loss, sensitivity to machine inefficiencies
(hT and hC), relatively lower cycle efficiency and costly fuel. Due to these factors,
the cost of power generation by a stationary gas turbine in a utility system is high.
However, a gas turbine plant offers certain advantages also, such as less installation
cost, less installation time, quick starting and stopping, and fast response to load
changes. So, a gas turbine plant is often used as a peaking unit for certain hours of
the day, when the energy demand is high. To utilize the high temperature exhaust
and to raise its plant efficiency a gas turbine may be used in conjunction with a
steam turbine plant to offer the gas turbine advantages of quick starting and stopping and permit flexible operation of the combined plant over a wide range of loads.
Let us consider two cyclic power plants coupled in series, the topping plant
operating on Brayton cycle and the bottoming one operating on Rankine cycle (Fig.
13.34). Helium may be the working fluid in the topping plant and water in the
bottoming plant. The overall efficiency of the combined plant is:
541
Gas Power Cycles
Q1
Q1
b
CC
C
GT
W1 = WT – WC
c
CC
W1
GT
C
Helium
a
HE
HE
d
Q2
1
Q2
B
P
T
W2
ST
W2 = WT – Wp
2
C
H2O
C
Q3
C.W.
4
c
(a)
Q1
b
p=
3
Q3
(b)
Brayton
cycle
c
p=
Helium
P
d
1
c
Q2
T
Rankine
cycle
a
H2O
4
3
Q3
2
s
(c)
Fig. 13.34 Brayton-Rankine Combined Cycle Plant
h = h1 + h2 – h1 h2
where h1 and h2 are the efficiencies of the Brayton cycle and Rankine cycle
respectively.
For capacity augmentation often supplementary firing is used (Fig. 13.35). For
expansion of combustion gases in the gas turbine
Tc/Td = (p2/p1)(g – 1)/g
where
g = 1.3 (assumed).
For the compressor, Tb/Ta = (p2/p1)(g ¢ – 1)/g ¢
where
g ¢ = 1.4.
WGT = wa [cpg (Tc – Td) – cpa (Tb – Ta)]
542
Engineering Thermodynamics
C
GT
c
b
a
1
d
Q2
h1
Supplementary
heating
Q4
Q1
e
ST
h2
Q5
2
c
Q6
HRSG
3
4
C.W.
f
(a)
c
e
d
b
1
f
T
a
4
3
2
s
(b)
Fig. 13.35
Brayton/Rankine Cyclic Plants with Supplementary Heating
neglecting the mass of fuel (for a high air-fuel ratio), and wa being the mass flow of
air.
WST = ws (h1 – h2)
where ws is the steam flow rate. The pump work is neglected. By energy balance,
wa cpg (Tc – Tf) = ws (h1 – h4)
Now,
Q1 = wa c pg [(Tc – Tb) + (Te – Td )]
The overall efficiency of the plant is:
W + WST
h = GT
Q1
Again,
Q1 = wf ¥ C.V.
where wf is the fuel burning rate.
543
Gas Power Cycles
High overall efficiency, low investment cost, less water requirement, large operating flexibility, phased installation, and low environmental impact are some of the
advantages of combined gas-steam cycles.
Solved Examples
Example 13.1 An engine working on the Otto cycle is supplied with air at
0.1 MPa, 35°C. The compression ratio is 8. Heat supplied is 2100 kJ/kg. Calculate
the maximum pressure and temperature of the cycle, the cycle efficiency, and the
mean effective pressure. (For air, cp = 1.005, cv = 0.718, and R = 0.287 kJ/kg K).
Solution
From Fig. 13.36
3
v
=
c
3
T
WE
2
1
Q2
=
2
c
Q1
v
P
pvg = c
4
4
Wc
1
s
v
(a)
(b)
Fig. 13.36
\
\
T1 = 273 + 35 = 308 K
p1 = 0.1 MPa = 100 kN/m2
Q1 = 2100 kJ/kg
rk = 8, g = 1.4
1
1
1
hcycle = 1 - g -1 = 1 - 0. 4 = 1 2.3
rk
8
= 0.565 or 56.5%
v1
RT1 0.287 ¥ 308
=
= 8, v1 =
= 0.844 m3/kg
p1
v2
100
0.884
v2 =
= 0.11 m3/kg
8
F I
GH JK
v1
T2
=
T1
v2
\
g -1
= (8)0.4 = 2.3
T2 = 2.3 ¥ 308 = 708.4 K
Q1 = cv (T3 – T2) = 2100 kJ/kg
544
\
or
Engineering Thermodynamics
2100
= 2925 K
0.718
T3 = Tmax = 3633 K
T3 – 708.4 =
g
F I = (8) = 18.37
GH JK
v1
p2
=
p1
v2
\
Again
p2 = 1.873 MPa
p3 v3
T3
\
1.4
=
p2 v2
T2
363
= 9.426 MPa
708
Wnet = Q1 ¥ hcycle = 2100 ¥ 0.565
= 1186.5 kJ/kg
= pm(v1 – v2) = pm(0.884 – 0.11)
p3 = pmax = 1.837 ¥
1186.5
= 1533 kPa = 1.533 MPa
0.774
Example 13.2 A Diesel engine has a compression ratio of 14 and cut-off takes
place at 6% of the stroke. Find the air standard efficiency.
Solution From Fig. 13.36
2
3
v
rk = 1 = 14
pvg = c
v2
v3 – v2 = 0.06 (v1 – v2)
= 0.06 (14 v2 – v2)
= 0.78 v2
4
v3 = 1.78 v2
1
v1
v
\
Cut-off ratio, rc =
= 1.78
v2
Fig. 13.36
pm = m.e.p. =
p
\
g
hDiesel = 1 -
1 1 rc - 1
.
.
g rkg -1 rc - 1
1. 4
1
1 (1.78) - 1
.
14 (14) 0. 4 1.78 - 1
1.24
= 1 – 0.248 ◊
= 0.605, i.e. 60.5%
0.78
Example 13.3 In an air standard Diesel cycle, the compression ratio is 16, and
at the beginning of isentropic compression, the temperature is 15°C and the pressure is 0.1 MPa. Heat is added until the temperature at the end of the constant
pressure process is 1480°C. Calculate (a) the cut-off ratio, (b) the heat supplied
per kg of air, (c) the cycle efficiency, and (d) the m.e.p.
= 1-
545
Gas Power Cycles
From Fig. 13.37
Solution
v1
= 16
v2
T1 = 273 + 15 = 288 K
p1 = 0.1 MPa = 100 kN/m2
T3 = 1480 + 273 = 1735 K
rk =
3
3
p
pvg = c
T
2
c
4
=
c
p=
v
2
4
1
1
v
s
(a)
(b)
Fig. 13.37
F I
GH JK
v1
T2
=
T1
v2
\
g -1
= (16)0.4 = 3.03
T2 = 288 ¥ 3.03 = 873 K
p2 v2
pv
= 3 3
T2
T3
T
v3
1753
= 3 =
= 2.01
T2
v2
273
(a) Cut-off ratio,
rc =
(b) Heat supplied,
Q1 = cp (T3 – T2)
= 1.005 (1753 – 873)
= 884.4 kJ/kg
F I
GH JK
v4
T3
=
v3
T4
g -1
Fv ¥v I
=G
H v v JK
1
2
2
3
g -1
=
F 16 I
H 2.01 K
0. 4
= 2.29
1753
= 766 K
2.29
Heat rejected,
Q2 = cv (T4 – T1) = 0.718 (766 – 288) = 343.2 kJ/kg
Q
(c)
Cycle efficiency = 1 – 2
Q1
\
T4 =
546
Engineering Thermodynamics
343.2
= 0.612 or 61.2%
884. 4
It may also be estimated from the equation
g
1 1 r -1
hcycle = 1 - ◊ g -1 ◊ c
g rk
rc - 1
=1–
1. 4
( 2.01) - 1
1
1
0 .4 ◊
1.4 (16)
2.01 - 1
1
1
◊
= 1◊ 1.64 = 0.612 or 61.2%
1.4 3.03
Wnet = Q1 ¥ hcycle
= 1-
= 884.4 ¥ 0.612 = 541.3 kJ/kg
RT1
0.287 ¥ 288
=
= 0.827 m3/kg
100
p1
v1 =
\
0.827
= 0.052 m3/kg
16
v1 – v2 = 0.827 – 0.052 = 0.775 m3/kg
(d)
m.e.p. =
v2 =
Wnet
541.3
=
= 698.45 kPa
v1 - v2
0.775
Example 13.4 An air standard dual cycle has a compression ratio of 16, and
compression begins at 1 bar, 50°C. The maximum pressure is 70 bar. The heat
transferred to air at constant pressure is equal to that at constant volume. Estimate
(a) the pressures and temperatures at the cardinal points of the cycle, (b) the cycle
efficiency, and (c) the m.e.p. of the cycle, cv = 0.718 kJ/kg K, cp = 1.005 kJ/kg K.
Given: (Fig. 13.38)
Solution
4
4
pvg = c
=c
p
p
T
3
3
2
2
5
1
v=
c
v=
c
1
v
s
(a)
(b)
Fig. 13.38
T1 = 273 + 50 = 323 K
F I
GH JK
v1
T2
=
v2
T1
g -1
= (16)0.4
5
547
Gas Power Cycles
\
T2 = 979 K
g
p2 = p1
F v I = 1.0 ¥ (16) = 48.5 bar
GH v JK
1.4
1
2
p3
70
= 979 ¥
= 1413 K
48.5
p2
Q2 – 3 = cv (T3 – T2) = 0.718 (1413 – 979) = 312 kJ/kg
T3 = T2 ◊
Now
\
\
Q2 – 3 = Q3 – 4 = cp (T4 – T3)
312
T4 =
+ 1413 = 1723 K
1.005
T
v4
1723
= 4 =
= 1.22
T3 1413
v3
v5
v
v
16
= 1 ¥ 3 =
= 13.1
v4
v2 v4 1.22
g -1
\
F v I = 1723 ¥ 1 = 615 K
T =T G J
(13.1)
Hv K
FT I
615
p = p G J = 1.0 ¥
= 1.9 bar
323
T
H K
4
5
4
0. 4
5
5
5
1
1
hcycle = 1 –
Q2
cv ( T5 - T1 )
=1–
cv ( T3 - T2 ) + c p ( T4 - T3 )
Q1
0.718(615 - 323)
312 + 312
0.718 ¥ 292
= 1= 0.665 or 66.5%
624
RT1
0.287kJ/ kg K ¥ 323K
v1 =
=
= 0.927 m3/kg
2
2
p1
10 kN / m
v
15
v1 – v2 = v1 – 1 =
v
16 16 1
Wnet = Q1 ¥ hcycle = 0.665 ¥ 624 kJ/kg
= 1-
\
Wnet
0.665 ¥ kJ/ kg
=
15
2
v1 - v2
¥ 0.927m / kg
16
= 476 kN/m2 = 4.76 bar
m.e.p. =
Example 13.5 In a gas turbine plant, working on the Brayton cycle with a regenerator of 75% effectiveness, the air at the inlet to the compressor is at
0.1 MPa, 30°C, the pressure ratio is 6, and the maximum cycle temperature is
900°C. If the turbine and compressor have each an efficiency of 80%, find the
percentage increase in the cycle efficiency due to regeneration.
548
Solution
Engineering Thermodynamics
Given: (Fig. 13.39)
p1 = 0.1 MPa
T1 = 303 K
T3 = 1173 K
rp = 6, hT = hC = 0.8
3
T
6
p
2
=
c
2s
4
p
=
c
4s
5
1
s
Fig. 13.39
Without a regenerator
T2 s
=
T1
Fp I
GH p JK
2
1
(g -1)/ g
=
T3
= (6)0.4/1.4 = 1.668
T4 s
T2s = 303 ¥ 1.668 = 505 K
1173
T4s =
= 705 K
1.668
T -T
505 - 303
T2 – T1 = 2 s 1 =
= 252 K
hC
0.8
T3 – T4 = hT (T3 – T4s) = 0.8 (1173 – 705) = 375 K
WT = h3 – h4 = cp (T3 – T4) = 1.005 ¥ 375 = 376.88 kJ/kg
WC = h2 – h1 = cp (T2 – T1) = 1.005 ¥ 252 = 253.26 kJ/kg
T2 = 252 + 303 = 555 K
Q1 = h3 – h2 = cp (T3 – T2) = 1.005 (1173 – 555) = 621.09 kJ/kg
W - WC
376.88 - 253.26
\
h= T
=
= 0.199 or 19.9%
621.09
Q1
With regenerator
T4 = T3 – 375 = 1173 – 375 = 798 K
T6 - T2
= 0.75
T4 - T2
T6 – 555 = 0.75 (798 – 555)
T6 = 737.3 K
Regenerator effectiveness =
\
or
549
Gas Power Cycles
\
Q1 = h3 – h6 = cp (T3 – T6)
= 1.005 (1173 – 737.3) = 437.88 kJ/kg
Wnet remains the same.
Wnet
123.62
=
= 0.2837 or 28.37%
437.9
Q1
\ Percentage increase due to regeneration
0.2837 - 0.199
=
= 0.4256, or 42.56%
0.199
Example 13.6 A gas turbine plant operates on the Brayton cycle between
Tmin = 300 K and Tmax = 1073 K. Find the maximum work done per kg of air, and
the corresponding cycle efficiency. How does this efficiency compare with the
Carnot cycle efficiency operating between the same two temperatures?
\
h=
Solution
(Wnet)max = cp ( Tmax - Tmin )
2
2
= 1.005 ( 1073 - 300 )
= 1.005 (15.43)2 = 239.28 kJ/kg
hcycle = 1 –
1
( rp )
( g -1)/ g
=1–
Tmin
Tmax
300
= 0.47 or 47%
1073
T
300
hCarnot = 1 – min = 1 –
= 0.721 or 72.1%
Tmax
1073
=1–
h Brayton
h Carnot
=
0.47
= 0.652
0.721
Example 13.7 In an ideal Brayton cycle, air from the atmosphere at 1 atm,
300 K is compressed to 6 atm and the maximum cycle temperature is limited to
1100 K by using a large air-fuel ratio. If the heat supply is 100 MW, find (a) the
thermal efficiency of the cycle, (b) work ratio, (c) power output, (d) exergy flow
rate of the exhaust gas leaving the turbine.
Solution The cycle efficiency,
1
1
hcycle = 1 – ( g -1)/ g = 1 - 0 . 4 /1. 4
6
rp
= 0.401 or 40.1%
T2/T1 = (rp)(g – 1)g = 1.67
T2 = 501 K
T3/T4 = 1.67, T4 = 1100/1.67 = 658.7 K
WC = 1.005 (501 – 300) = 202 kJ/kg
WT = 1.005 (1100 – 658.7) = 443.5 kJ/kg
550
Engineering Thermodynamics
WT - WC
241.5
=
= 0.545
443.5
WT
Power output = 100 ¥ 0.401 = 40.1 MW
Work ratio =
.
Q1 = m cp (T3 – T2) = 100,000 kW
.
m = 166.1 kg/s
Exergy flow rate of the exhaust gas stream
T
T
.
= m cp T0 4 - 1 - ln 4
T0
T0
F
GH
= 166.1 ¥ 1.005 ¥ 300
I
JK
F 658 - 1 - ln 658.7 I
H 300
300 K
= 20.53 MW
Example 13.8 The following refer to a stationary gas turbine:
Compressor inlet temperature = 311 K
Compressor pressure ratio = 8
Combustion chamber pressure drop = 5% of inlet pressure
Turbine inlet temperature = 1367 K
Turbine exit and compressor inlet pressures are atmospheric.
There exists a facility to take air from the compressor exit for use in cooling the
turbine. Find the percentage of air that may be taken from the compressor for this
purpose so that the overall cycle efficiency drops by 5% from that of the case of no
usage of compressed air for cooling of turbine. For simplicity, assume the following: (a) Take properties of gas through the turbine as those of air, (b) Addition of
cooling air to the turbine and addition of fuel to the combustion chamber do not
affect the turbine power, (c) Compressor and turbine efficiencies are 0.87 and 0.90
respectively.
Solution Given:
hC = 0.87, hT = 0.9, T1 = 311 K, p2/p1 = 8,
p3 = 0.95p2, T3 = 1367 K, p4 = p1 = 1 atm, g = 1.4
Fig. 13.40
551
Gas Power Cycles
Case-1: No cooling
WC =
.
m a c pa ( T2s - T1 )
hc
.
WT = m g cpg (T3 – T4s ) hT
.
Q1 = m cpc (T3 – T2)
.
.
.
.
ma = mg = mc = m
Here,
cpa = cpg = cpc = cp
(T2s/T1 = p2s/p1)(g – 1)/g = 80.4/1.4 = 1.181
T2s = 563.3 K
T2 s - T1
= 0.87, T2 = 601 K
T2 - T1
(g – 1)/g
T3/T4s = (p3/p4s )
F 0.95 p I
=G
H p JK
2
0 . 4 /1. 4
= 1.785
1
T4s = 765.83 K
.
.
.
WC = 290 m cp, WT = 541.06 m cp and Q1 = 766 m cp
hcycle =
541.06 - 290
= 0.328
766
Case-2: With cooling
hcycle = 0.328 – 0.05 = 0.278
Since the extraction of compressed air for turbine cooling does not contribute to
turbine work or burner fuel flow, it can be treated as an increment x added to the
compressor mass flow.
541.06 - 290(1 + x )
= 0.278
766
\
x = 0.13
0.13
% of compressor delivery air flow =
¥ 100 = 11.6%
113
.
Example 13.9 In a gas turbine plant the ratio of Tmax/Tmin is fixed. Two
arrangements of components are to be investigated: (a) single-stage compression
followed by expansion in two turbines of equal pressure ratios with reheat to the
maximum cycle temperature, and (b) compression in two compressors of equal
pressure ratios, with intercooling to the minimum cycle temperature, followed by
single-stage expansion. If hC and hT are the compressor and turbine efficiencies,
show that the optimum specific output is obtained at the same overall pressure
ratio for each arrangement.
If hC is 0.85 and hT is 0.9, and Tmax /Tmin is 3.5, determine the above pressure
ratio for optimum specific output and show that with arrangement (a) the optimum
output exceeds that of arrangement (b) by about 11%.
Solution
(a) With reference to Fig. 13.41 (a)
552
Engineering Thermodynamics
CC
CC
2
3
C
T1
1
4
5
L
T2
Wc
6
3
Tmax
5
T
2
4
2s
6
4s
6s
Tmin
1
s
Fig. 13.41(a)
p2
p
= 4
p4
p1
T1 = Tmin, T3 = T5 = Tmax,
\
p4 =
p1 p2
p2 s
= r, pressure ratio
p1
\
p2s = p2 = rp1
\
p4 =
T2 s
=
T1
where
\
x=
r ◊ p1
FG p IJ b g = r
HpK
g -1 / g
2
x
1
g -1
g
T2s = Tmin rx
(DTs)comp = T2s – T1 = Tmin r x – Tmin = Tmin (r x – 1)
\
(DT )comp =
d
i
Tmin r x - 1
hC
FG IJ b g = FG rp IJ r
H K
H r ◊p K
p3
T3
=
T4 s
p4
\
T4s = T3 r
x
g -1 / g
–x/2
1
1
= Tmax r
–x/2
x/2
553
Gas Power Cycles
(DTs)turb = T3 – T4s = Tmax – Tmax ◊ r–x/2
= Tmax (1 – r–x/2 )
(DT)turb 1 = hT Tmax (1 – r– x/2 ) = (DT)turb 2
\
LM
d
N
L
dW
x
= c M2h T
◊r
2
dr
N
i Th dr - 1iOPQ
O
T
x ◊r P = 0
h
Q
Wnet = cp 2h T Tmax 1 - r - x / 2 -
\
net
p
min
x
C
- x / 2 -1
T max
min
x -1
C
On simplification
Tmax
Tmin
r3x/2 = hT hC
F T IJ
r = Gh h
H T K
b g
2 g / 3 g -1
max
\
opt
T
C
min
(b) With reference to Fig. 13.41 (b)
T2 s
=
Tmin
FG p IJ b g = d r i = r
Hp K
g -1 / g
x
2
x/2
1
3
5
CC
2
4
C1
C2
L
T
6
1
5
T
Tmax
4s
4
3
2
6
2s
1
6s
Tmin
s
Fig. 13.41 (b)
(DTs)comp 1 = T2s – T1 = Tmin (rx/2 – 1)
(DT)comp1 =
Tmin(r x/2 −1)
ηC
= (DT )comp 2
554
Engineering Thermodynamics
FG IJ b g = r
H K
Tmax
p5
=
p6
T6s
g -1 / g
x
T6s = Tmax ◊ g –x
\
(DTs)turb = Tmax – T6s = Tmax (1 – r–x )
(DT)turb = hT Tmax (1 – r–x )
L
2T dr - 1i O
PP
W = c Mh T d1 - r i h
MN
Q
L
dW
2T
x a f O
◊r
= c Mh T
x◊r a f PQ = 0
h
dr
2
N
x /2
min
-x
net
p
T max
C
- x +1
net
p
min
x / 2 -1
T max
C
On simplification
FG
H
ropt = h T h C
Tmax
Tmin
IJ
K
b g
2g / 3 g -1
This is the same as in (a).
If
hC = 0.85, hT = 0.9
1
g
1.4 7
=
=
=
g - 1 0.4 2
x
Tmax
= 3.5
Tmin
\
ropt = (0.85 ¥ 0.9 ¥ 3.5)2/3 ¥ 7/2 = 9.933
LM
O
T
r - 1iP
d
i
d
h
N
Q
= c LM2 ¥ 0.9 T d1 - 9.933
i - 0T.85 d9.933 - 1iOPQ
N
L
O
F 1 I - T 1178
= c ◊ T M2 ¥ 0.9 ¥ 3.5 1 H 1.388K . a0.928fPQ
N
Wnet (a) = cp 2h T Tmax 1 - r - x / 2 -
min
x
C
-0.143
p
p
min
0.286
max
min
= 0.670 cp ◊ Tmin
LM
N
d
i 0.285 d9.933
Wnet (b) = c p Tmin 0.9 ¥ 3.5 1 - 9.933-0 . 286 -
0 .143
iOPQ
-1
= cp Tmin (1.518 – 0.914)
= 0.604 c p Tmin
af
af
Wnet a - Wnet b
¥ 100
Wnet a
=
af
0.670 - 0.604
¥ 100 = 10.9%
0.670
Proved.
555
Gas Power Cycles
Example 13.10 A turbojet aircraft flies with a velocity of 300 m/s at an altitude
where the air is at 0.35 bar and – 40ºC. The compressor has a pressure ratio of 10,
and the temperature of the gases at the turbine inlet is 1100ºC. Air enters the compressor at a rate of 50 kg/s. Estimate (a) the temperature and pressure of the gases
at the turbine exit, (b) the velocity of gases at the nozzle exit, and (c) the propulsive
efficiency of the cycle.
Solution
(a) For isentropic flow of air in the diffusor (Fig. 13.42)
=0
=0
V 22 – V 21
=0
Q1–2 – W1–2
= h2 – h1 +
0 = cp (T2 – T1) –
T2 = T1 +
2
V12
2
V12
300 2
= 233 +
¥ 10–3
2 ¥ 1.005
2c p
= 277.78 K
p2 = p1 (T2/T1)g/(g – 1)
F
H
kN 277.78
m2
233
= 35
I
K
1.4 / 0 .4
= 64.76 kPa
p3 = rp p2 = 10 ¥ 64.76 = 647.6 kPa
F p I b g T = 277.78 (10)
T =G J
Hp K
g -1 / g
3
3
0.4/1.4
2
2
= 536.66 K
WC = W T
h3 – h2 = h4 – h5
or,
T3 – T2 = T4 – T5
T5 = T4 – T3 + T2 = 1373 – 536.66 + 277.78
= 1114.12 K
b g
F
T I
F 1114.12 I
p =G J
p = 647.6
H 1373 K
HT K
g / g -1
5
5
4
4
= 311.69 K
3.5
556
Engineering Thermodynamics
Fig. 13.42
(b) For isentropic expansion of gases in the nozzle,
F p I b g = 1114.12 F 35 I
GH p JK
H 311.69K
g -1 / g
T6 = T5
0.286
6
5
= 596.12 K
Neglecting the K.E. of gas at nozzle inlet,
V6 = [2cp (T5 – T6) ¥ 1000]1/2
= [2 ¥ 1.005 (1114.12 – 596.12) ¥ 1000]1/2
= 1020.4 m/s
(c) The propulsive efficiency of a turbojet engine is the ratio of the propulsive
.
power developed Wp to the total heat transfer to the fluid
.
WP = w[Vexit – Vinlet ] Vaircraft
= 50 [1020.4 – 300] ¥ 300
kg m 2
¥ 2
s
s
= 10.806 MW
Q1 = w (h4 – h3) = 50 ¥ 1.005 (1373 – 536.66)
= 42.026 MW
hP =
10.806
= 0.257 or 25.7%
42.026
Example 13.11 In a combined GT — ST plant, the exhaust gas from the GT is
the supply gas to the steam generator at which a further supply of fuel is burned in
the gas. The pressure ratio for the GT is 8, the inlet air temperature is 15ºC and the
maximum cycle temperature is 800ºC.
Gas Power Cycles
557
Combustion in the steam generator raises the gas temperature to 800ºC and the
gas leaves the generator at 100ºC. The steam condition at supply is at 60 bar,
600ºC and the condenser pressure is 0.05 bar. Calculate the flow rates of air and
steam required for a total power output of 190 MW and the overall efficiency of the
combined plant. What would be the air-fuel ratio? Assume ideal processes. Take
cpg = 1.11 and cpa = 1.005 kJ/kg K, gg = 1.33, ga = 1.4, C.V. of fuel = 43.3 MJ/kg.
Neglect the effect of fuel flow on the total mass flow of gas expanding in the gas
turbine.
Solution
With reference to Fig. 13.35.
Tb = Ta (pb/pa](g – 1)/g = 288 ¥ 80.4/1.4 = 522 K
Td =
TC
1073
1073
= 638 K
= 0.331.33 =
1.682
rp g - 1 / g 8
b g
WGT = cp (Tc – Td) – cp (Tb – Ta)
a
g
= 1.11 (1073 – 638) – 1.005 (522 – 288)
= 249 kJ/kg
Q1 = cp (Tc – Tb) = 1.11 (1073 – 522)
g
= 612 kJ/kg
Q¢1 = cpg (Te – Td) = 1.11(1073 – 638)
= 483 kJ/kg
h1 = 3775, h2 = 2183, h3 = 138 = h4, all in kJ/kg
(Q1)St = h1 – h3 = 3775 – 138 = 3637 kJ/kg
Q¢¢fe = 1.11 (800 – 100) = 777 kJ/kg
By energy balance of the steam generator,
wa ¥ 777 = ws ¥ 3637
wa/ws = 4.68
WST = h1 – h2 = 3775 – 2183 = 1592 kJ/kg
wa ¥ 249 + ws ¥ 1592 = 190 ¥ 103 kW
ws (4.68 ¥ 249 + 1592) = ws ¥ 2.757 ¥ 103 = 190 ¥ 103
ws = 68.9 kg/s and wa = 322.5 kg/s
Now,
wa(612 + 483) = wf ¥ 43,300
wa/wf = A/F ratio =
Fuel energy input =
43,300
= 39.5
1095
322.5
¥ 43,300 = 353525 kW
39.5
= 353.5 MW
h0A =
190
= 0.537 or 53.7%
353.5
558
Engineering Thermodynamics
Review Questions
13.1 What are cyclic and non-cyclic heat engines? Give examples.
13.2 What are the four processes which constitute the Stirling cycle? Show that the
regenerative Stirling cycle has the same efficiency as the Carnot cycle.
13.3 State the four processes that constitute the Ericsson cycle. Show that the
regenerative Ericsson cycle has the same efficiency as the Carnot cycle.
13.4 Mention the merits and demerits of the Stirling and Ericsson cycles.
13.5 What is an air standard cycle? Why are such cycles conceived?
13.6 What is a spark ignition engine? What is the air standard cycle of such an
engine? What are its four processes?
13.7 Show that the efficiency of the Otto cycle depends only on the compression
ratio.
13.8 How is the compression ratio of an SI engine fixed?
13.9 What is a compression ignition engine? Why is the compression ratio of such
an engine more than that of an SI engine?
13.10 State the four processes of the Diesel cycle.
13.11 Explain the mixed or dual cycle.
13.12 For the same compression ratio and heat rejection, which cycle is most efficient: Otto, Diesel or Dual? Explain with p–v and T–s diagrams.
13.13 With the help of p–v and T–s diagrams, show that for the same maximum pressure and temperature of the cycle and the same heat rejection,
hDiesel > hDual > hOtto
13.14 What are the three basic components of a gas turbine plant? What is the air
standard cycle of such a plant? What are the processes it consists of?
13.15 Show that the efficiency of the Brayton cycle depends only on the pressure
ratio.
13.16 What is the application of the closed cycle gas turbine plant?
13.17 How does Brayton cycle compare with Rankine cycle?
13.18 Discuss the merits and demerits of Brayton and Otto cycles applied to reciprocating and rotating plants.
13.19 What is the effect of regeneration on Brayton cycle efficiency? Define the
effectiveness of a regenerator.
13.20 What is the effect of irreversibilities in turbine and compressor on Brayton
cycle efficiency?
13.21 Explain the effect of pressure ratio on the net output and efficiency of a Brayton
cycle.
13.22 Drive the expression of optimum pressure ratio for maximum net work output
in an ideal Brayton cycle. What is the corresponding cycle efficiency?
13.23 Explain the effects of: (a) intercooling, and (b) reheating, on Brayton cycle.
13.24 What is a free shaft turbine?
13.25 With the help of flow and T–s diagrams explain the air standard cycle for a jet
propulsion plant.
13.26 Define propulsive power and propulsive efficiency.
13.27 Why are regenerators and intercoolers not used in aircraft engines? What is
after burning? Why is it used?
Gas Power Cycles
559
13.28 Explain the working of a turbofan engine with the help of a neat sketch. Define
“bypass ratio”. How does it influence the engine thrust?
13.29 How does a turboprop engine differ from a turbofan engine?
13.30 What is a ramjet? How is the thrust produced here?
13.31 What is a rocket? How is it propelled?
13.32 Explain the advantages and disadvantages of a gas turbine plant for a utility
system.
13.33 What are the advantages of a combined gas turbine-steam turbine power plant?
13.34 With the help of flow and T–s diagrams explain the operation of a combined
GT–ST plant. Why is supplementary firing often used?
Problems
13.1 In a Stirling cycle the volume varies between 0.03 and 0.06 m3, the maximum
pressure is 0.2 MPa, and the temperature varies between 540ºC and 270ºC. The
working fluid is air (an ideal gas). (a) Find the efficiency and the work done per
cycle for the simple cycle. (b) Find the efficiency and the work done per cycle
for the cycle with an ideal regenerator, and compare with the Carnot cycle having the same isothermal heat supply process and the same temperature range.
Ans. (a) 27.7%, 53.7 kJ/kg, (b) 32.2%
13.2 An Ericsson cycle operating with an ideal regenerator works between 1100 K
and 288 K. The pressure at the beginning of isothermal compression is 1.013
bar. Determine (a) the compressor and turbine work per kg of air, and (b) the
cycle efficiency.
Ans. (a) WT = 465 kJ/kg, WC = 121.8 kJ/kg (b) 0.738
13.3 Plot the efficiency of the air standard Otto cycle as a function of the compression ratio for compression ratios from 4 to 16.
13.4 Find the air standard efficiencies for Otto cycles with a compression ratio of 6
using ideal gases having specific heat ratios 1.3, 1.4 and 1.67. What are the
advantages and disadvantages of using helium as the working fluid?
13.5 An engine equipped with a cylinder having a bore of 15 cm and a stroke of
45 cm operates on an Otto cycle. If the clearance volume is 2000 cm3, compute
the air standard efficiency.
Ans. 47.4%
13.6 In an air standard Otto cycle the compression ratio is 7, and compression
begins at 35ºC, 0.1 MPa. The maximum temperature of the cycle is 1100ºC.
Find (a) the temperature and pressure at the cardinal points of the cycle, (b) the
heat supplied per kg of air, (c) the work done per kg of air, (d) the cycle efficiency, and (e) the m.e.p. of the cycle.
13.7 An engine working on the Otto cycle has an air standard cycle efficiency of
56% and rejects 544 kJ/kg of air. The pressure and temperature of air at the
beginning of compression are 0.1 MPa and 60ºC respectively. Compute (a) the
compression ratio of the engine, (b) the work done per kg of air, (c) the pressure and temperature at the end of compression, and (d) the maximum pressure
in the cycle.
13.8 For an air standard Diesel cycle with a compression ratio of 15 plot the
efficiency as a function of the cut-off ratio for cut-off ratios from 1 to 4. Compare with the results of Problem 13.3.
560
Engineering Thermodynamics
13.9 In an air standard Diesel cycle, the compression ratio is 15. Compression begins at 0.1 MPa, 40ºC. The heat added is 1.675 MJ/kg. Find (a) the maximum
temperature of the cycle, (b) the work done per kg of air, (c) the cycle efficiency, (d) the temperature at the end of the isentropic expansion, (e) the cutoff ratio, (f) the maximum pressure of the cycle, and (g) the m.e.p. of the cycle.
13.10 Two engines are to operate on Otto and Diesel cycles with the following data:
Maximum temperature 1400 K, exhaust temperature 700 K. State of air at the
beginning of compression 0.1 MPa, 300 K.
Estimate the compression ratios, the maximum pressures, efficiencies, and rate
of work outputs (for 1 kg/min of air) of the respective cycles.
Ans. Otto—rk = 5.656, pmax = 2.64 MPa, W = 2872 kJ/kg, h = 50%
Diesel—rk = 7.456, pmax = 1.665 MPa, W = 446.45 kJ/kg, h = 60.8%
13.11 An air standard limited pressure cycle has a compression ratio of 15 and
compression begins at 0.1 MPa, 40ºC. The maximum pressure is limited to
6 MPa and the heat added is 1.675 MJ/kg. Compute (a) the heat supplied at
constant volume per kg of air, (b) the heat supplied at constant pressure per kg
of air, (c) the work done per kg of air, (d) the cycle efficiency, (e) the temperature at the end of the constant volume heating process, (f) the cut-off ratio,
and (g) the m.e.p. of the cycle.
Ans. (a) 235 kJ/kg, (b) 1440 kJ/kg, (c) 1014 kJ/kg,
(d) 60.5%, (e) 1252 K, (f) 2.144 (g) 1.21 MPa
13.12 In an ideal cycle for an internal combustion engine the pressure and temperature at the beginning of adiabatic compression are respectively 0.11 MPa and
115°C, the compression ratio being 16. At the end of compression heat is added
to the workign fluid, first, at constant volume, and then at constant pressure
reversibly. The working fluid is then expanded adiabatically and reversibly to
the original volume.
If the working fluid is air and the maximum pressure and temperature are
respectively 6 MPa and 2000°C, determine, per kg of air (a) the pressure,
temperature, volume, and entropy of the air at the five cardinal points of the
cycle (take s1 as the entropy of air at the beginning of compression), and (b) the
work output and efficiency of the cycle.
13.13 Show that the air standard efficiency for a cycle comprising two constant pressure processes and two isothermal processes (all reversible) is given by
eg -1jg
T1 - T2 i In e rp j
d
h=
L
eg -1j /g - T O
T1 M1 + In e rp j
2P
N
Q
where T1 and T2 are the maximum and minimum temperatures of the cycle, and
rp is the pressure ratio.
13.14 Obtain an expression for the specific work done by an engine working on the
Otto cycle in terms of the maximum and minimum temperatures of the cycle,
the compression ratio rk, and constants of the working fluid (assumed to be an
ideal gas).
Hence show that the compression ratio for maximum specific work output is
given by
FG T IJ b g
HT K
1/ 2 1- g
rk =
min
max
Gas Power Cycles
561
13.15 A dual combustion cycle operates with a volumetric compression ratio rk = 12,
and with a cut-off ratio 1.615. The maximum pressure is given by pmax = 54p1,
where p1 is the pressure before compression. Assuming indices of compression
and expansion of 1.35, show that the m.e.p. of the cycle
pm = 10 p1
Hence evaluate (a) temperatures at cardinal points with T1 = 335 K, and (b)
cycle efficiency.
Ans. (a) T2 = 805 K, p2 = 29.2 p1, T3 = 1490 K,
T4 = 2410 K, T5 = 1200 K, (b) h = 0.67
13.16 Recalculate (a) the temperatures at the cardinal points, (b) the m.e.p., and (c)
the cycle efficiency when the cycle of Problem 13.15 is a Diesel cycle with the
same compression ratio and a cut-off ratio such as to give an expansion curve
coincident with the lower part of that of the dual cycle of Problem 13.15.
Ans. (a) T2 = 805 K, T3 = 1970 K, T4 = 1142 K (b) 6.82 p1, (c) h = 0.513
13.17 In an air standard Brayton cycle the compression ratio is 7 and the maximum
temperature of the cycle is 800ºC. The compression begins at 0.1 MPa, 35ºC.
Compare the maximum specific volume and the maximum pressure with the
Otto cycle of Problem 13.6. Find (a) the heat supplied per kg of air, (b) the net
work done per kg of air, (c) the cycle efficiency, and (d) the temperature at the
end of the expansion process.
13.18 A gas turbine plant operates on the Brayton cycle between the temperatures
27ºC and 800ºC. (a) Find the pressure ratio at which the cycle efficiency
approaches the Carnot cycle efficiency, (b) find the pressure ratio at which the
work done per kg of air is maximum, and (c) compare the efficiency at this
pressure ratio with the Carnot efficiency for the given temperatures.
13.19 In a gas turbine plant working on the Brayton cycle the air at the inlet is at
27ºC, 0.1 MPa. The pressure ratio is 6.25 and the maximum temperature is
800ºC. The turbine and compressor efficiencies are each 80%. Find (a) the
compressor work per kg of air, (b) the turbine work per kg of air, (c) the heat
supplied per kg of air, (d) the cycle efficiency, and (e) the turbine exhaust temperature.
Ans. (a) 259.4 kJ/kg, (b) 351.68 kJ/kg, (c) 569.43 kJ/kg,
(d) 16.2%, (e) 723 K
13.20 Solve Problem 13.19 if a regenerator of 75% effectiveness is added to the plant.
13.21 Solve Problem 13.19 if the compression is divided into two stages, each of
pressure ratio 2.5 and efficiency 80%, with intercooling to 27ºC.
13.22 Solve Problem 13.21 if a regenerator of 75% effectiveness is added to the plant.
13.23 Solve Problem 13.19 if a reheat cycle is used. The turbine expansion is divided
into two stages, each of pressure ratio 2.5 and efficiency 80%, with reheat to
800ºC.
13.24 Solve Problem 13.23 if a regenerator of 75% effectiveness is added to the plant.
13.25 Solve Problem 13.24 if the staged compression of Problem 13.21 is used in the
plant.
13.26 Find the inlet condition for the free-shaft turbine if a two-shaft arrangement is
used in Problem 13.19.
13.27 A simple gas turbine plant operating on the Brayton cycle has air inlet temperature 27ºC, pressure ratio 9, and maximum cycle temperature 727ºC.
What will be the improvement in cycle efficiency and output if the turbine
process is divided into two stages each of pressure ratio 3, with intermediate
reheating to 727ºC?
Ans. – 18.3%, 30.6%
562
Engineering Thermodynamics
13.28 Obtain an expression for the specific work output of a gas turbine unit in terms
of pressure ratio, isentropic efficiencies of the compressor and turbine, and the
maximum and minimum temperatures, T3 and T1. Hence show that the pressure
ratio rp for maximum power is given by
FG
H
rp = η T η C
T3
T1
IJ b g
K
γ / 2 γ −1
If T3 = 1073 K, T1 = 300 K, hC = 0.8, hT = 0.8 and g = 1.4 compute the optimum
value of pressure ratio, the maximum net work output per kg of air, and corresponding cycle efficiency.
Ans. 4.263, 100 kJ/kg, 17.2%
13.29 A gas turbine plant draws in air at 1.013 bar, 10ºC and has a pressure ratio of
5.5. The maximum temperature in the cycle is limited to 750ºC.
Compression is conducted in an uncooled rotary compressor having an isentropic efficiency of 82%, and expansion takes place in a turbine with an isentropic efficiency of 85%. A heat exchanger with an efficiency of 70% is fitted
between the compressor outlet and combustion chamber. For an air flow of
40 kg/s, find (a) the overall cycle efficiency, (b) the turbine output, and (c) the
air-fuel ratio if the calorific value of the fuel used is 45.22 MJ/kg.
Ans. (a) 30.4%, (b) 4272 kW, (c) 115
13.30 A gas turbine for use as an automotive engine is shown in Fig. 13.43. In the
first turbine, the gas expands to just a low enough pressure p5, for the turbine to
drive the compressor. The gas is then expanded through a second turbine connected to the drive wheels. Consider air as the working fluid, and assume that
all processes are ideal. Determine (a) pressure p5, (b) the net work per kg and
mass flow rate, (c) temperature T3 and cycle thermal efficiency, and (d) the T-s
diagram for the cycle.
Regenerator
Exahaust
7
6
3
2
CC
5
4
C
Wnet = 150 kW
T
Power
Turbine
p2
p1 = 4.0
t4 = 920°C
p7 = 1 atm
Wc
Air intake
1
p1 = 1 atm
t1 = 25°C
Fig. 13.43
13.31 Repeat Problem 13.30 assuming that the compressor has an efficiency of 80%,
both the turbines have efficiencies of 85%, and the regenerator has an efficiency of 72%.
13.32 An ideal air cycle consists of isentropic compression, constant volume heat
transfer, isothermal expansion to the original pressure, and constant pressure
heat transfer to the original temperature. Deduce an expression for the cycle
efficiency in terms of volumetric compression ratio rk , and isothermal
563
Gas Power Cycles
expansion ratio, rk. In such a cycle, the pressure and temperature at the start of
compression are 1 bar and 40ºC, the compression ratio is 8, and the maximum
pressure is 100 bar. Determine the cycle efficiency and the m.e.p.
Ans. 51.5%, 3.45 bar
13.33 For a gas turbine jet propulsion unit, shown in Fig. 13.28, the pressure and
temperature entering the compressor are 1 atm and 15ºC respectively. The pressure ratio across the compressor is 6 to 1 and the temperature at the turbine
inlet is 1000ºC. On leaving the turbine the air enters the nozzle and expands to
1 atm. Determine the pressure at the nozzle inlet and the velocity of the air
leaving the nozzle.
13.34 Repeat Problem 13.33, assuming that the efficiency of the compressor and turbine are both 85%, and that the nozzle efficiency is 95%.
13.35 Develop expressions for work output per kg and the efficiency of an ideal
Brayton cycle with regeneration, assuming maximum possible regeneration.
For fixed maximum and minimum temperatures, how do the efficiency and
work outputs vary with the pressure ratio? What is the optimum pressure ratio?
13.36 For an air standard Otto cycle with fixed intake and maximum temperatures, T1
and T3, find the compression ratio that renders the net work per cycle a
maximum. Derive the expression for cycle efficiency at this compression ratio.
If the air intake temperature, T1, is 300 K and the maximum cycle temperature, T3, is 1200 K, compute the compression ratio for maximum net work,
maximum work output per kg in a cycle, and the corresponding cycle efficiency.
Find the changes in work output and cycle efficiency when the compression
ratio is increased from this optimum value to 8. Take cv = 0.718 kJ/kg K.
Ans. 5.65, 215 kJ/kg, 50%, DWnet = – 9 kJ/kg, Dh = + 6.4%
13.37 Show that the mean effective pressure, pm, for the Otto cycle is given by
e p - p r j FGH1 - r 1 IJK
p =
bg - 1gbr - 1g
3
g
1k
g -1
k
m
k
where p3 = pmax, p1 = pmin and rk is the compression ratio.
13.38 A gas turbine plant operates on the Brayton cycle using an optimum pressure
ratio for maximum net work output and a regenerator of 100% effectiveness.
Derive expressions for net work output per kg of air and corresponding
efficiency of the cycle in terms of the maximum and the minimum temperatures.
If the maximum and minimum temperatures are 800ºC and 30ºC respectively,
compute the optimum value of pressure ratio, the maximum net work output
per kg and the corresponding cycle efficiency.
Ans. (Wnet)max = cp
dT
max -
Tmin
i
2
Tmin
, (rp)opt = 9.14,
Tmax
(Wnet)max = 236.79 kJ/kg; hcycle = 0.469
13.39 A gas turbine unit is to provide peaking power for an electrical utility with a net
power output of 10 MW. The pressure ratio across the compressor is 7, the
efficiency of the compressor 80%, and the efficiency of the turbine is 92%. In
(hcycle)max = 1 –
564
Engineering Thermodynamics
order to conserve fuel, a regenerator with an effectiveness of 85% is used. The
maximum temperature of the cycle is 1200 K. The air at compressor inlet is at
20ºC, 1.1 bar. Assume the working fluid to be air which behaves as an ideal gas
with cp = 1.005 kJ/kg K and g = 1.4. Neglect pressure drops in the combustion
chamber and regenerator. Determine the required air flow and the fuel flow
rates for a fuel heating value of 42 MJ/kg, and the power plant efficiency.
13.40 Show that for the Stirling cycle with all the processes occurring reversibly but
where the heat rejected is not used for regenerative heating, the efficiency is
given by
T1
- 1 + g - 1 ln r
T2
h = 1T1
T
- 1 + g - 1 1 ln r
T2
T2
FG
H
FG
H
IJ b g
K
IJ b g
K
where r is the compression ratio and T1/T2 the maximum to minimum temperature ratio.
Determine the efficiency of this cycle using hydrogen (R = 4.307 kJ/kg K,
cp = 14.50 kJ/kg K) with a pressure and temperature prior to isothermal
compression of 1 bar and 300 K respectively, a maximum pressure of 2.55 MPa
and heat supplied during the constant volume heating of 9300 kJ/kg. If the heat
rejected during the constant volume cooling can be utilized to provide the
constant volume heating, what will be the cycle efficiency? Without altering
the temperature ratio, can the efficiency be further improved in the cycle?
13.41 Helium is used as the working fluid in an ideal Brayton cycle. Gas enters the
compressor at 27ºC and 20 bar and is discharged at 60 bar. The gas is heated to
1000ºC before entering the turbine. The cooler returns the hot turbine exhaust
to the temperature of the compressor inlet. Determine: (a) the temperatures at
the end of compression and expansion, (b) the heat supplied, the heat rejected
and the net work per kg of He, and (c) the cycle efficiency and the heat rate.
Take cp = 5.1926 kJ/kg K.
Ans. (a) 4 65.5, 820.2 K, (b) 4192.5, 2701.2, 1491.3 kJ/kg,
(c) 0.3557, 10,121 kJ/kWh
13.42 In an air standard cycle for a gas turbine jet propulsion unit, the pressure and
temperature entering the compressor are 100 kPa and 290 K, respectively. The
pressure ratio across the compressor is 6 to 1 and the temperature at the turbine
inlet is 1400 K. On leaving the turbine the air enters the nozzle and expands to
100 kPa. Assuming that the efficiency of the compressor and turbine are both
85% and that the nozzle efficiency is 95%, determine the pressure at the nozzle
inlet and the velocity of the air leaving the nozzle.
Ans. 285 kPa, 760 m/s
13.43 A stationary gas turbine power plant operates on the Brayton cycle and delivers
20 MW to an electric generator. The maximum temperature is 1200 K and the
minimum temperature is 290 K. The minimum pressure is 95 kPa and the maximum pressure is 380 kPa. If the isentropic efficiencies of the turbine and compressor are 0.85 and 0.80 respectively, find (a) the mass flow rate of air to the
compressor, (b) the volume flow rate of air to the compressor, (c) the fraction
of the turbine work output needed to drive the compressor, (d) the cycle efficiency.
Gas Power Cycles
565
If a regenerator of 75% effectiveness is added to the plant, what would be the
changes in the cycle efficiency and the net work output?
Ans. (a) 126.37 kg/s, (b) 110.71 m3/s, (c) 0.528,
(d) 0.2146, Dh = 0.148 DWnet = 0
13.44 Air enters the compressor of a gas turbine operating on Brayton cycle at 1 bar,
27ºC. The pressure ratio in the cycle is 6. Calculate the maximum temperature
in the cycle and the cycle efficiency. Assume WT = 2.5 WC and g = 1.4.
Ans. 1251.4 K, 40%
13.45 In an ideal jet propulsion cycle, air enters the compressor at 1 atm, 15°C. The
pressure of air leaving the compressor is 5 atm and the maximum temperature
is 870°C. The air expands in the turbine to such a pressure that the turbine work
is equal to the compressor work. On leaving the turbine the air expands reversibly and adiabatically in a nozzle to 1 atm. Determine the velocity of air leaving
the nozzle.
Ans. 710.3 m/s
13.46 In a gas turbine the compressor is driven by the h.p. turbine. The exhaust from
the h.p. turbine goes to a free-shaft l.p. turbine which runs the load. The air
flow rate is 20 kg/s, and the minimum and maximum temperatures are respectively 300 K and 1000 K. The compressor pressure ratio is 4. Calculate the
pressure ratio of the l.p. turbine and the temperature of the exhaust gases from
the unit. The compressor and turbine are isentropic. Take c p of air and exhaust
gases as 1 kJ/kg K and g = 1.4.
Ans. 2.3, 673 K
13.47 A regenerative gas turbine with intercooling and reheat operates at steady state.
Air enters the compressor at 100 kPa, 300 K with a mass flow rate of 5.807 kg/s.
The pressure ratio across the two-stage compressor as well as the turbine is 10.
The intercooler and reheater each operate at 300 kPa. At the inlets to the turbine stages, the temperature is 1400 K. The temperature at inlet to the second
compressor stage is 300 K. The efficiency of each compressor and turbine stage
is 80%. The regenerator effectiveness is 80%. Determine (a) the thermal efficiency, (b) the back work ratio, WC /WT, (c) the net power developed.
Ans. (a) 0.443, (b) 0.454, (c) 2046 kW
13.48 In a regenerative gas turbine power plant air enters the compressor at 1 bar,
27ºC and is compressed to 4 bar. The isentropic efficiency of the compressor is
80% and the regenerator effectiveness is 90%. All of the power developed by
the h.p. turbine is used to drive the compressor and the l.p. turbine provides the
net power output of 97 kW. Each turbine has an isentropic efficiency of 87%
and the temperature at inlet to the h.p. turbine is 1200 K. Determine (a) the
mass flow rate of air into the compressor, (b) the thermal efficiency, (c) the
temperature of the air at the exit of the regenerator.
Ans. (a) 0.562 kg/s, (b) 0.432, (c) 523.2 K
14
Refrigeration Cycles
14.1 REFRIGERATION BY NON-CYCLIC PROCESSES
Refrigeration is the cooling of a system below the temperature of its surroundings.
The melting of ice or snow was one of the earliest methods of refrigeration and
is still employed. Ice melts at 0°C. So when ice is placed in a given space warmer
than 0°C, heat flows into the ice and the space is cooled or refrigerated. The latent
heat of fusion of ice is supplied from the surroundings, and the ice changes its state
from solid to liquid.
Another medium of refrigeration is solid carbon dioxide or dry ice. At atmospheric pressure CO2 cannot exist in a liquid state, and consequently, when solid
CO2 is exposed to atmosphere, it sublimates, i.e. it goes directly from solid to
vapour, by absorbing the latent heat of sublimation (620 kJ/kg at 1 atm, –78.5°C)
from the surroundings (Fig. 14.1). Thus dry ice is suitable for low temperature
refrigeration.
T
pcr = 73 atm
tcr = 31ºC
S
– 60ºC
– 78.5ºC
S
+
L
V
L
L+V
ptr = 5.11 atm
ttr = – 60ºC
Heat of
sublimation
1 atm
S+V
620 kJ/kg
s
Fig. 14.1 T-s Diagram of CO2
In these two examples it is observed that the refrigeration effect has been
accomplished by non-cyclic processes. Of greater importance, however, are the
methods in which the cooling substance is not consumed and discarded, but used
again and again in a thermodynamic cycle.
567
Refrigeration Cycles
14.2 REVERSED HEAT ENGINE CYCLE
A reversed heat engine cycle, as explained in Sec. 6.12, is visualized as an engine
operating in the reverse way, i.e. receiving heat from a low temperature region,
discharging heat to a high temperature region, and receiving a net inflow of work (Fig. 14.2). Under such
T1
conditions the cycle is called a heat pump cycle or a
refrigeration cycle (see Sec. 6.6). For a heat pump
Q =Q +W
1
2
Q1
Q
(COP)H.P. = 1 =
W
W
Q1 - Q2
and for a refrigerator
Q2
Q2
Q
(COP)ref = 2 =
Q1 - Q2
W
T2
The working fluid in a refrigeration cycle is called
a refrigerant. In the reversed Carnot cycle (Fig. 14.3), Fig. 14.2 Reversed Heat
the refrigerant is first compressed reversibly and adiaEngine Cycle
batically in process 1-2 where the work input per kg
of refrigerant is W c, then it is condensed reversibly in
process 2-3 where the heat rejection is Q1, the refrigerant then expands reversibly
and adiabatically in process 3-4 where the work output is W E, and finally it absorbs
heat Q 2 reversibly by evaporation from the surroundings in process 4-1.
Q1
Condenser
3
Compressor
WC
T
Q1
2
T1
3 p1
Expansion
engine
2
WE
p2
4 Evaporator
T2
1
4
WE
WC
Q2
Q2
1
s
(a)
(b)
Fig. 14.3
Reversed Carnot Cycle
Here
Q1 = T1 (s2 – s3 ), Q2 = T2 (s1 – s4)
and
Wnet = W C – WE = Q 1 – Q2 = (T1 – T2) (s1 – s4)
where T1 is the temperature of heat rejection and T2 the temperature of heat
absorption.
Q
T2
(COP ref)rev = 2 =
Wnet
T1 - T2
and
(COP H.P.)rev =
Q1
T1
=
Wnet
T1 - T2
(14.1)
568
Engineering Thermodynamics
As shown in Sec. 6.16, these are the maximum values for any refrigerator or heat
pump operating between T1 and T2. It is important to note that for the same T2 or T1,
the COP increases with the decrease in the temperature difference (T1 – T2), i.e. the
closer the temperatures T1 and T2, the higher the COP.
14.3
VAPOUR COMPRESSION REFRIGERATION CYCLE
In an actual vapour refrigeration cycle, an expansion engine, as shown in
Fig. 14.3, is not used, since power recovery is small and does not justify the
cost of the engine. A throttling valve or a capillary tube is used for expansion in
reducing the pressure from p1 to p2. The basic operations involved in a vapour
compression refrigeration plant are illustrated in the flow diagram, Fig. 14.4, and
the property diagrams, Fig. 14.5.
Q1
2
3
p1
Condenser
Wc
p2
Expansion
valve
1
Compressor
4
Evaporator
Q2
Fig. 14.4 Vapour Compression Refrigeration Plant-flow Diagram
p1 2¢
2
1
1¢
4
p2
1¢
1
3
p1
p2
4
1
1¢
s
v
(a)
2¢
2¢
h =c
c
c
s=
h=
p2
4
p1
3
2
h
3
T
p
2
(b)
s
(c)
Fig. 14.5 Vapour Compression Refrigeration Cycle-Property Diagrams
The operations represented are as follows for an idealized plant:
1. Compression A reversible adiabatic process 1–2 or 1¢–2¢ either starting with
saturated vapour (state 1), called dry compression, or starting with wet vapour
(state 1¢), called wet compression. Dry compression (1-2) is always preferred to
569
Refrigeration Cycles
wet compression (1¢-2¢), because with wet compression there is a danger of the
liquid refrigerant being trapped in the head of the cylinder by the rising piston
which may damage the valves or the cylinder head, and the droplets of liquid refrigerant may wash away the lubricating oil from the walls of the cylinder, thus accelerating wear.
2. Cooling and Condensing A reversible constant pressure process, 2-3, first
desuperheated and then condensed, ending with saturated liquid. Heat Q1 is transferred out.
3. Expansion An adiabatic throttling process 3-4, for which enthalpy remains
unchanged. States 3 and 4 are equilibrium points. Process 3-4 is adiabatic (then
only h3 = h4 by S.F.E.E.), but not isentropic.
p2 vdp
Tds = dh – vdp, or s4 – s3 = –
p1 T
Hence it is irreversible and cannot be shown in property diagrams. States 3 and 4
have simply been joined by a dotted line.
z
4. Evaporation A constant pressure reversible process, 4-1, which completes
the cycle. The refrigerant is throttled by the expansion valve to a pressure, the
saturation temperature at this pressure being below the temperature of the surroundings. Heat then flows, by virtue of temperature difference, from the surroundings,
which gets cooled or refrigerated, to the refrigerant, which then evaporates, absorbing the latent heat of evaporation. The evaporator thus produces the cooling or the
refrigerating effect, absorbing heat Q2 from the surroundings by evaporation.
In refrigeration practice, enthalpy is the most sought-after property. The diagram in p-h coordinates is found to be the most convenient. The constant property
lines in the p-h diagram are shown in Fig. 14.6, and the vapour compression cycle
in Fig. 14.7.
Critical point
c
x=c
x=
x=
c
s=c
s=c
s=c
c
Liquid
p
t3
x=
t2
t1
t3
t2
t1 t increasing
Sat liq.
line
(x = 0)
Liquid vap.
mixture
s=c
v=c
Vapour
v=c
v=c
h
Sat.
Vap.
line
(x = 1)
Fig. 14.6 Phase Diagram with Constant Property Lines on p-h Plot
570
Engineering Thermodynamics
p
Q1
p1
3
2
Wc
Q2
s=
X
4
p2
c
t 2 = Compressor
discharge
temperature
1
4
h
Fig. 14.7
Vapour Compression Cycle on p-h Diagram
14.3.1 Performance and Capacity of a Vapour Compression
Plant
Figure 14.8 shows the simplified diagram of a vapour compression refrigeration
plant.
Q1
3
Condenser
Space to be cooled
Expansion
or refrigerated
valve
2
Wc
4
1
Evaporator
Compressor
Q2
Fig. 14.8 Vapour Compression Plant
When steady state has been reached, for 1 kg flow of refrigerant through the
cycle, the steady flow energy equations (neglecting K.E. and P.E. changes) may be
written for each of the components in the cycle as given below.
Compressor
h1 + W c = h2
\
Wc = (h2 – h1) kJ/kg
Condenser
h2 = Q1 + h3
\
Q1 = (h2 – h3) kJ/kg
Expansion valve
h3 = h4
or
\
(hf )p1 = (hf )p2 + x4 (hfg)p2
x4 =
( h f ) p1 - ( h f ) p2
( h f ) p2
This is the quality of the refrigerant at the inlet to the evaporator (mass fraction of
vapour in liquid-vapour mixture).
Refrigeration Cycles
571
Evaporator
h4 = Q2 = h1
\
Q2 = (h1 – h4) kJ/kg
This is known as the refrigerating effect, the amount of heat removed from the
surroundings per unit mass flow of refrigerant.
If the p-h chart for a particular refrigerant is available with the given parameters,
it is possible to obtain from the chart the values of enthalpy at all the cardinal points
of the cycle. Then for the cycle
Q
h - h4
COP = 2 = 1
(14.2)
Wc
h2 - h1
If w is the mass flow of refrigerant in kg/s, then the rate of heat removal from the
surroundings
= w(h1 – h4) kJ/s = w (h1 – h4) ¥ 3600 kJ/h
One tonne of refrigeration is defined as the rate of heat removal from the surroundings equivalent to the heat required for melting 1 tonne of ice in one day. If the
latent heat of fusion of ice is taken as 336 kJ/kg, then 1 tonne is equivalent to heat
revoval at the rate of (1000 ¥ 336)/24 or 14,000 kJ/h
\
Capacity of the refrigerating plant
w (h1 - h4 ) ¥ 3600
tonnes
14,000
The rate of heat removal in the condenser
Q1 = w(h2 – h3) kJ/s
� c the flow-rate of cooling water in kg/s, and (tc2
If the condenser is water-cooled, m
– tc1) the rise in temperature of water, then
� c cc (tc2 – tc1) kJ/s
Q1 = w(h2 – h3) = m
provided the heat transfer is confined only between the refrigerant and water, and
there is no heat interaction with the surroundings.
The rate of work input to the compressor
Wc = w(h2 – h1) kJ/s
14.3.2
Actual Vapour Compression Cycle
In order to ascertain that there is no droplet of liquid refrigerant being carried over
into the compressor, some superheating of vapour is recommended after the evaporator.
A small degree of subcooling of the liquid refrigerant after the condenser is also
used to reduce the mass of vapour formed during expansion, so that too many vapour
bubbles do not impede the flow of liquid refrigerant through the expansion valve.
Both the superheating of vapour at the evaporator outlet and the subcooling of
liquid at the condenser outlet contribute to an increase in the refrigerating effect, as
shown in Fig. 14.9. The compressor discharge temperature, however, increases,
due to superheat, from t ¢2 to t2, and the load on the condenser also increases.
572
Engineering Thermodynamics
Degree of subcooling
c.p.
t2
3
3
2
t2
p
2
1
x4
4
4
x4
1
t1
t1
Degree of superheat
h
Fig. 14.9 Superheat and Subcooling in a Vapour Compression Cycle
Sometimes, a liquid-line heat exchanger is used in the plant, as shown in
Fig. 14.10. The liquid is subcooled in the heat exchanger, reducing the load on the
condenser and improving the COP. For 1 kg flow
Q2 = h6 – h5, Q 1 = h2 – h3
Wc = h2 – h1 and h1 – h6 = h3 – h4
14.3.3
Components in a Vapour Compression Plant
Condenser It must desuperheat and then condense the compressed refrigerant.
Condensers may be either air-cooled or water-cooled. An air-cooled condenser is
used in small self-contained units. Water-cooled condensers are used in larger installations.
Q1
Condenser
p
3
Suction line
heat exchanger
1
4
Expansion
valve
3
4
2
2
6
Wc
Compressor
5
6
h
5 Evaporator
(a)
1
(b)
Fig. 14.10 Vapour Compression Cycle with a Suction-line Heat Exchanger
Expansion device
It reduces the pressure of the refrigerant, and also regulates
the flow of the refrigerant to the evaporator. Two widely used types of expansion
devices are: capillary tubes and throttle valves (thermostatic expansion valves).
Capillary tubes are used only for small units. Once the size and length are fixed, the
evaporator pressure, etc. gets fixed. No modification in operating conditions is possible. Throttle valves are used in larger units. These regulate the flow of the refrigerant according to the load on the evaporator.
Compressor
Compressors may be of three types: (1) reciprocating, (b) rotary,
and (c) centrifugal. When the volume flow rate of the refrigerant is large,
Refrigeration Cycles
573
centrifugal compressors are used. Rotary compressors are used for small units.
Reciprocating compressors are used in plants up to 100 tonnes capacity. For plants
of higher capacities, centrifugal compressors are employed.
In reciprocating compressors, which may be single-cylinder or multi-cylinder
ones, because of clearance, leakage past the piston and valves, and throttling
effects at the suction and discharge valves, the actual volume of gas drawn into the
cylinder is less than the volume displaced by the piston. This is accounted for in the
term volumetric efficiency, which is defined as
Actual volume of gas drawn at evaporator pressure and temperature
hvol =
Piston displacement
\ Volume of gas handled by the compressor
p 2 N
= w ◊ v1(m3/s) =
D L n ¥ h vol
4
60
where w is the refrigerant flow rate,
v1 is the specific volume of the refrigerant at the compressor inlet,
D and L are the diameter and stroke of the compressor,
n is the number of cylinders in the compressor, and
N is the r.p.m.
The clearance volumetric efficiency is given by Eq. (18.13)
F
H
I
K
h vol = 1 + C – C
FG p IJ
Hp K
1/ n
2
1
where C is the clearance.
Evaporator A common type of evaporator is a coil brazed on to a plate, called a
plate evaporator. In a ‘flooded evaporator’ the coil is filled only with a liquid refrigerant. In an indirect expansion coil, water (up to 0°C) or brine (for temperatures
between 0 and – 21°C) may be chilled in the evaporator, and the chilled water or
brine may then be used to cool some other medium.
14.3.4
Multistage Vapour Compression Systems
For a given condensation temperature, the lower the evaporator temperature, the
higher becomes the compressor pressure ratio. For a reciprocating compressor, a
high pressure ratio across a single stage means low volumetric efficiency. Also,
with dry compression the high pressure ratio results in high compressor discharge
temperature which may damage the refrigerant. To reduce the work of compression
and improve the COP, multistage compression with intercooling may
be adopted. Since the intercooler temperature may be below the temperature of
available cooling water used for the condenser, the refrigerant itself may be used as
the intercooling medium. Figure 14.11 shows a two-stage compression system with
a direct contact heat exchanger.
As shown in Sec. 18.4, for minimum work, the intercooler pressure pi is the
geometric mean of the evaporator and condenser pressures, p1 and p2, or
pi =
p1 ◊ p2
574
Engineering Thermodynamics
Q1
Condenser
4
5
m1
3
C.V.
6
2
m2
Direct
contact heat
exchanger
H.P.
Compressor
1
7
Evaporator
8
L.P. Compressor
Q2
(a)
p2
4
p
5
p1
7
3
6
m1
2
m2
8
p1
1
h
(b)
Fig. 14.11 Two-stage Vapour Compression System
By making an energy balance of the direct contact heat exchanger,
� 2h2 + m
� 1h6 = m
� 2 h7 + m
� 1 h3
m
m� 1
h - h7
= 2
h3 - h6
m� 2
The desired refrigerating effect determines the flow rate in the low pressure loop,
� 2, as given below
m
14000
� 2 (h1 – h8) =
m
¥P
3600
where P is the capacity, in tonnes of refrigeration.
3.89 P
�2=
m
\
kg/s
h1 - h8
Figure 14.12 shows a two-stage vapour compression system with a flash chamber
intercooler, where the vapour from the flash chamber (state 9) mixes with the vapour
from the LP compressor (state 2) to form vapour at state 3, which enters the HP
compressor.
\
575
Refrigeration Cycles
Q1
5
4
Condenser
H.P.
Compressor
3
9
2
6
Wc2
Flash
chamber
L.P.
Compressor
7
1
Evaporator
8
Q2
Wc1
(a)
4
p
5
9
7
3
6
2
1
8
h
(b)
Fig. 14.12 Two-stage Vapour Compression System with a Flash Intercooler
14.3.5
Refrigerants
The most widely used refrigerants now-a-days are a group of halogenated
hydrocarbons, marketed under the various proprietary names of freon, genetron,
arcton, isotron, frigen, etc. These are either methane-based or ethane-based, where
the hydrogen atoms are replaced by chlorine or fluorine atoms. Methane-based compounds are denoted by a number of two digits, where the first digit minus one is the
number of hydrogen atoms and the second digit indicates the number of fluorine
atoms, while the other atoms are chlorine. For Refrigerant-12 (R–12), e.g. the number of hydrogen atoms is zero, the number of fluorine atoms is two, and hence the
other two atoms must be chlorine. Therefore, the compound is CCl2F2, dichlorodifluoro methane. Similarly, for R-22, it is CHClF2, monochloro-difluoro methane;
for R-50, it is methane, CH4; for R-10, it is CCl4, carbon tetrachloride, and so on. If
the compound is ethane-based, a three-digit number is assigned to the refrigerant,
where the first digit is always 1, the second digit minus one is the number of hydro-
576
Engineering Thermodynamics
gen atoms, and the third digit indicates the number of fluorine atoms, all other atoms
in the hydrocarbon being chlorine. For example, R-110 is C2Cl6, R-113 is C2Cl3F3,
R-142 is C2H3ClF2, and so on. The use of these refrigerants is now discouraged,
since these, being largely insoluble in water, move up, react with ozone in the ozone
layer (which protects the earth from pernicious ultraviolet rays) and deplete it.
It was realized in mid-seventies that the CFCs (chloro-fluorocarbons) not only
allow more ultraviolet radiation into the earth’s atmosphere, but also prevent the
infrared radiation from escaping the earth to outer space, which contributes to the
greenhouse effect and hence, global warming. As a result, the use of some CFCs is
banned (by Montreal Protocol, 1987) and phased out in many countries. Fully halogenated CFCs (such as R-11, R-12 and R-115) do the most damage to the ozone
layer. The partially halogenated refrigerants such as R-22 have about 5% of the
ozone depleting potential (ODP) of R-12. CFCs, friendly to the ozone layer that
protects the earth from ultraviolet rays and which do not contribute to the greenhouse effect are being developed. The chlorine free R-134a, a recent finding, is
presently replacing R-12, the most widely used refrigerant, particularly in domestic
refrigerators and freezers and automotive air conditioners.
Two important parameters that need to be considered in the selection of a
refrigerant are the temperatures of the two media (the refrigerated space and the
environment), with which the refrigerant exchanges heat. To have reasonable heat
transfer rate, a temperature difference of 5 to 10°C should be maintained between
the refrigerant and the medium. If a space is to be maintained at – 10°C, e.g. the
refrigerant should evaporate at about – 20°C (Fig. 14.13), the saturation pressure at
which should be above atmospheric pressure to prevent any air leakage into the
system. Again, the temperature of the refrigerant in the condenser should be above
the cooling medium by about 10°C, as shown in the figure, the saturation pressure
at which must be below
the critical pressure of the
refrigerant. If a single re2
frigerant cannot meet the
temperature requirements
(–20°C to 50°C range),
50ºC
two cycles with two dif3
10ºC
ferent refrigerants can be
T
40ºC
used in series (Fig. 14.14).
Such a coupled cycle
makes a cascade refrig- C
– 10ºC
eration system.
10ºC
Other desirable charac4
1
– 20ºC
teristics of a refrigerant
are that it should be nons
toxic, noncorrosive, nonFig. 14.13
flammable and chemically
stable, should have a large enthalpy of vaporization to minimize the mass flow, and
should be available at low cost.
577
Refrigeration Cycles
Q1
40ºC
High-temperature
condenser
7
6
50ºC
Expansion
valve
Intermediate
heat exchange
8
WB
Compressor
Cycle B
5
3
2
Cycle A
–10ºC
4
WA
Compressor
Expansion
valve
1
Low-temperature
evaporator
–20ºC
Q2
(a)
Q1
50ºC
6
7
p
40ºC
Temp.
overlap
WB
Cycle B
2
3
5
8
Cycle A
WA
10ºC
4
Q2
–20ºC
1
h
(b)
Fig. 14.14
Cascade Refrigeration Cycle
Ammonia is widely used in food refrigeration facilities such as the cooling of
fresh fruits, vegetables, meat and fish, refrigeration of beverages and dairy products such as beer, wine, milk and cheese, freezing of ice cream and ice production,
low temperature refrigeration in the pharmaceutical and other process industries.
The advantages of ammonia are its low cost, higher COPs and thus lower energy
578
Engineering Thermodynamics
costs, greater detectability in the event of a leak, no effect on the ozone layer, and
more favourable thermodynamic and transport properties and thus higher heat transfer coefficients requiring smaller and lower cost heat exchangers. The major drawback of ammonia is its toxicity which makes it unsuitable for domestic use.
Other fluids used as refrigerants are sulphur dioxide, methyl chloride, ethyl chloride, hydrocarbons like propane, butane, ethane, ethylene, etc. carbon dioxide, air
and water.
14.4
ABSORPTION REFRIGERATION CYCLE
The absorption refrigeration system is a heat operated unit which uses a refrigerant
that is alternately absorbed and liberated from the absorbent. In the basic absorption system, the compressor in the vapour compression cycle is replaced by an
absorber-generator assembly involving less mechanical work. Figure 14.15 gives
the basic absorption refrigeration cycle, in which ammonia is the refrigerant and
water is the absorbent. This is known as the aqua-ammonia absorption system.
NH3 Vapour
QG
QC
Condenser
Steam or
electricity
Generator
Cooling water
NH3 Liquid
Heat
exchanger
Wp
Aqua-pump
Strong solution
Expansion
valve
Reducing valve
QE
Weak solution
NH3 Vapour
Absorber
QA
Evaporator
Cooling water
Brine
(NaCl or CaCl2
in water)
Fig. 14.15 Vapour Absorption Refrigeration Plant-flow Diagram
Ammonia vapour is vigorously absorbed in water. So when low-pressure ammonia vapour from the evaporator comes in contact in the absorber with the weak
solution (the concentration of ammonia in water is low) coming from the generator,
it is readily absorbed, releasing the latent heat of condensation. The temperature of
the solution tends to rise, while the absorber is cooled by the circulating water,
absorbing the heat of solution (Q A), and maintaining a constant temperature. Strong
solution, rich in ammonia, is pumped to the generator where heat (QG) is supplied
from an external source (steam, electricity, gas flame, etc.). Since the boiling point
of ammonia is less than that of water, the ammonia vapour is given off from the
aqua-ammonia solution at high pressure, and the weak solution returns to the absorber through a pressure reducing valve. The heat exchanger preheats the strong
579
Refrigeration Cycles
solution and precools the weak solution, reducing both QG and QA, the heat to be
supplied in the generator and the heat to be removed in the absorber respectively.
The ammonia vapour then condenses in the condenser, is throttled by the expansion
valve, and then evaporates, absorbing the heat of evaporation from the surroundings or the brine to be chilled.
In driving the ammonia vapour out of the solution in the generator, it is impossible to avoid evaporating some of the water. This water vapour going to the
condenser along with the ammonia vapour, after condensation, may get frozen
to ice and block the expansion valve. So an analyzer-rectifier combination
(Fig. 14.16) is used to eliminate water vapour from the ammonia vapour going into
the condenser.
Rectifier
NH3 Vapour
Qc
Analyser
Drip
Condenser
Cooling water
QG
Cooling water
Suction line
heat exchanger
Heating
medium
Heat
exchanger
Generator
Weak
solution
Expansion
valve
NH3
Vapour
Reducing
valve
QE
Medium
to be cooled
QA
Strong
solution
Fig. 14.16
Cooling
water
Evaporator
Absorber
Aqua pump
Actual Vapour Absorption Refrigeration Plant with Analyzer and Rectifier
The analyzer is a direct-contact heat exchanger consisting of a series of trays
mounted above the generator. The strong solution from the absorber flows downward over the trays to cool the outgoing vapours. Since the saturation temperature
of water is higher than that of ammonia at a given pressure, it is the water vapour
which condenses first. As the vapour passes upward through the analyzer, it is
cooled and enriched by ammonia, and the liquid is heated. Thus the vapour going to
the condenser is lower in temperature and richer in ammonia, and the heat input to
the generator is decreased.
The final reduction in the percentage of water vapour in the ammonia going to
the condenser occurs in the rectifier which is a water-cooled heat exchanger which
condenses water vapour and returns it to the generator through the drip line, as
shown in Fig. 14.16. The use of a suction-line heat exchanger is to reduce Q A and
increase Q E, thus achieving a double benefit. In the absorber the weak solution is
580
Engineering Thermodynamics
sprayed to expose a larger surface area so as to accelerate the rate of absorption of
ammonia vapour.
There is another absorption refrigeration system, namely, lithium bromide-water
vapour absorption (Fig. 14.17). Here the refrigerant is water and the absorbent is
the solution of lithium bromide salt in water. Since water cannot be cooled below
0°C, it can be used as a refrigerant in air conditioning units. Lithium bromide solution has a strong affinity for water vapour because of its very low vapour pressure.
It absorbs water vapour as fast as it is released in the evaporator.
H 2O Vapour
QC
105∞C
7.5 cm Hg abs.
105∞C
Q G Generator
Weak
solution
65%
Heat
exchanger
63∞C, Strong
solution
Cooling water
Steam or
any heat
source
Condenser
H 2O Vapour
Heat
exchanger
7.5 cm Hg abs.
0.7 cm Hg
QE
0.7 cm Hg
QA
60% conc.
Absorber Cooling
water
Solution pump
Fig. 14.17
7∞C
Refrigeration
load
Evaporator Water 13∞C
Pump
Lithium Bromide-water Absorption Refrigeration Plant
While the vapour compression refrigeration system requires the expenditure
of ‘high-grade’ energy in the form of shaft work to drive the compressor with the
concommitant disadvantage of vibration and noise, the absorption refrigeration system requires only ‘low-grade’ energy in the form of heat to drive it, and
it is relatively silent in operation and subject to little wear. Although the COP =
QE/QG is low, the absorption units are usually built when waste heat is available,
and they are built in relatively bigger sizes. One current application of the absorption system that may grow in importance is the utilization of solar energy for the
generator heat source of a refrigerator for food preservation and perhaps for comfort cooling.
14.4.1
Theoretical COP of an Absorption System
Let us assume that an absorption refrigeration plant uses heat QG from a source at
T1, provides refrigeration Q E for a region at TR, and rejects heat (QA + Q C) to a sink
(atmosphere) at T2, as shown in Fig. 14.18.
By the first low
QE + QG = QC + QA
(14.3)
581
Refrigeration Cycles
By the second law
(DS)source + (DS)sink + (DS)region ≥ 0
\
–
QG QE + QG QE
≥0
+
T1
T2
TR
Source, T 1
Absorption unit
QG
Generator
QC
Pump
Condenser
Q A Sink, T 2
W
(Negligible)
Absorber
Evaporator
QE
Region, T R
Energy Fluxes in Vapour Absorption Plant
Fig. 14.18
From Eq. (14.3)
–
QG QE + QG QE
+
≥0
T1
T2
TR
\
T1 - T2
T - T2
QG + R
QE ≥ 0
T1T2
T2 TR
\
T2 - TR
T - T2
QE £ 1
QG
T2 TR
T1T2
\
QE
(T - T ) T
£ 1 2 R
QG
(T2 - TR ) T1
or
COP £
(T1 - T2 ) TR
(T2 - TR ) T1
\
(COP)max =
( T1 - T2 ) TR
( T2 - TR ) T1
or
(COP)max =
TR
T - T2
¥ 1
T2 - TR
T1
(14.4)
Therefore, the maximum possible COP is the product of the ideal COP of a refrigerator working between TR and T2, and the ideal thermal efficiency of an engine
working between T1 and T2. The cyclic heat engine and the refrigerator together, as
shown in Fig. 14.19, is equivalent to the absorption cycle.
582
14.5
Engineering Thermodynamics
HEAT PUMP SYSTEM
The heat pump is a cyclic device
T 1 (Generator)
which is able to extract energy at
QG
a low temperature heat source and
Cyclic heat engine
upgrade it to a high temperature
W
heat source, enabling it to be used
QA
T 2 (Condenser
more effectively. Low grade reand Absorber)
QC
ject heat available at a low temperature may be upgraded to a
W
Cyclic refrigerator
high temperature heat source by a
heat pump. While a refrigerator is
QE
meant for the removal of heat and
T R (Evaporator)
to achieve cooling, a heat pump is
used to supply heat at a high Fig. 14.19 An Absorption Cycle as Equivalent
to a Cyclic Heat Engine and a
temperature.
Cyclic Refrigerator
Heat pump systems have many
features in common with the refrigeration systems and may be of the vapour-compression or the absorption type. A typical vapour-compression heat pump for space
heating (Fig. 14.20) has the same basic components as the vapour-compression
refrigeration system: compressor, condenser, expansion valve, and evaporator. The
heat Q� in comes from the surroundings (cold air), and Q� out is directed to the dwelling as the desired effect, with the COP given by:
Q�
h - h3
COP = out = 2
h2 - h1
WC
Many possible sources are available for heat transfer ( Q� in) to the refrigerant
passing through the evaporator. These include the outside air, the ground, and water
from lakes, rivers, or wells. Liquid circulated through a solar collector and stored in
an insulated tank also can be used as a source for a heat pump. Industrial heat
pumps employ waste heat or warm liquid or gas streams as the low temperature heat
source and upgrade it.
Inside air
Condenser
Q out
3
Evaporator
4
Expansion
valve
Wc
2
Q in
Compressor
1
Outside
air
Fig. 14.20 Vapour-Compression Heat Pump System for Space Heating
583
Refrigeration Cycles
An air-air heat pump can be used for year-round air conditioning, to achieve
heating during winter and cooling during summer, by using a reversing valve with
the evaporator and condenser executing opposite duties. A heat pump in industry
will be very effective if both condenser and evaporator are utilized for heating
and cooling respectively, e.g. for cooling a process stream and heating another
(Fig. 14.21).
Feed 1
Feed 2
Hot
Cold
Compressor
Evaporator
Condenser
Cold
Expansion
valve
Hot
Fig. 14.21 Heating and Cooling of Two Process Streams
14.6
GAS CYCLE REFRIGERATION
Refrigeration can also be accomplished by means of a gas cycle. In the gas cycle, an
expander replaces the throttle valve of a vapour compression system, because the
drop in temperature by throttling a real gas is very small. For an ideal gas, enthalpy
is a function of temperature only, and since in throttling enthalpy remains unchanged, there would not be any change in temperature also. Work output obtained
from the expander is used as an aid in compression, thus decreasing the net work
input. The ideal gas-refrigeration cycle is the same as the reversed Brayton cycle.
The flow and T-s diagrams of the cycle are shown in Fig. 14.22. Since there is no
phase change, the condenser and evaporator in a vapour compression system are
here called the cooler and refrigerator respectively. The COP of the refrigeration
cycle, assuming the gas to be ideal, is given by
Q
h1 - h4
COP = 2 =
Wnet
( h2 - h1 ) - (h3 - h4 )
=
T1 - T4
=
(T2 - T1 ) - (T3 - T4 )
T1 - T4
T1
FG T - 1IJ - T FG T - 1IJ
H T K HT K
4
1
For isentropic compression and expansion
FG IJ
H K
T2
p
= 1
T1
p2
\
( g -1)/ g
=
COP =
T3
T4
T4
T1 - T4
=
T3 - T4
T
(T1 - T4 ) 3 - 1
T4
FG
H
IJ
K
3
2
4
584
Engineering Thermodynamics
Also
COP =
1
FG p IJ
Hp K
(14.5)
(g -1)/ g
1
-1
2
where p1 is the pressure after compression and p2 is the pressure before compression.
2
Q1
T
p1
3
p1
Cooler
Q2
2
3
Q2
p1
Compressor
WE
p2
4
1
p2
M
Expander
WC
Q1
4
s
Refrigerator
Fig. 14.22 Gas Refrigeration Cycle
The COP of a gas-cycle refrigeration system is low. The power requirement per
unit capacity is high. Its prominent application is in aircrafts and missiles, where
the vapour compression refrigeration system becomes heavy and bulky. Figure
14.23 shows the open cycle aircraft cabin cooling.
The compressed air is available and is a small percentage of the amount handled
by the compressor of a turbojet or a supercharged aircraft engine. Large amounts of
cool ambient air are available for cooling the compressed air. In addition to cooling, the replacement of stale air in the cabin is possible. At high altitudes the pressurization of cabin air is also possible. Because of these considerations, air cycle
refrigeration is favoured in aircrafts.
2 p1
Bleed from main
power plant
compressor
2
3
4
p1
T
Ambient
air
3
Air turbine
p2
Air cabin
p2
Bypass control
4
s
Fig. 14.23
14.7
Open Cycle Aircraft Cabin Cooling
LIQUEFACTION OF GASES
An important application of gas refrigeration processes is in the liquefaction of
gases. A gas may be cooled either by making it expand isentropically in an
expander, thus performing work (sometimes known as external-work method), or
585
Refrigeration Cycles
making the gas undergo Joule-Kelvin expansion through a throttle valve (sometimes called the internal-work method). While the former method always brings
about a temperature decrease, the expansion through the throttle valve may yield a
temperature decrease only when the temperature before throttling is below the maximum inversion temperature (see Sec. 11.7).
14.7.1 Linde-Hampson System for Liquefaction of Air
In this system, Joule-Kelvin effect is utilized for cooling, and ultimately, liquefying
the air. The schematic diagram and the T-s diagram are shown in Fig. 14.24. Ideally, the compression would be isothermal as shown on the T-s diagram. A twostage compressor with intercooling and aftercooling is shown. The yield, Y, of the
system is defined as the ratio of the mass of liquid produced to the mass of gas
compressed. The energy required per unit mass of liquid produced is known as the
specific work consumption, W.
Cooling
water
Cooling water
L.P. Compressor
After cooler
Intercooler
Purifier
C2
C1
H.P. Compressor
m - m1
Expansion valve
Joule-Kelvin
refrigeration system
2
1
8
7
Heat
exchanger
m
M
Make-up
.
mf
Control volume
3
6
.
m
4
Separator
5
m1
Liquid yield, Y
(a)
8
2
1
h=c
1a
tm
T
20
0a
tm
7
3
4
5
6
s
(b)
Fig. 14.24
Linde-Hampson Cycle for Air Liquefaction
586
Engineering Thermodynamics
The theoretical yield, assuming perfect insulation, can be determined by a mass
and energy balance for the control volume (Joule-Kelvin refrigeration system) as
� f be the rate at which liquid air is produced (the same
shown in Fig. 14.24. Let m
� the rate at which air is compressed and
given to the system as make-up), and m
then expanded. Then the yield is
m� f
Y=
m�
and the energy balance gives
� h2 – m
� f h5 – ( m
� –m
� f )h7 = 0
m
\
� (h2 – h7) – m
� f (h5 – h7) = 0
m
m� f
h - h7
= 2
m�
h5 - h7
h - h2
\
Y= 7
(14.6)
h7 - h5
No yield is thus possible unless h7 is greater than h2. The energy balance for the
compressor gives
� h1 + Wc = m
� h2 + Q R
m
where QR is the heat loss to the surroundings from the compressor
Wc
\
= T1 (s1 – s2) – (h1 – h2)
m�
This is the minimum work requirement.
Specific work consumption, W
W
W 1 h - h5
m�
= c ¥
= c = 7
[T1 (s1 – s2) – (h1 – h2)]
m�
m� f
m� Y h7 - h2
Y=
14.7.2
Claude System of Air Liquefaction
In the Claude system, energy is removed from the gas stream by allowing it to do
some work in an expander. The flow and T-s diagrams are given in Fig. 14.25.
The gas is first compressed to pressures of about 40 atm and then passed through
the first heat exchanger. Approximatly 80% of the gas is then diverted from the
main stream, expanded through an expander, and reunited with the return stream
below the second heat exchanger. The stream to be liquefied continues through the
second and third heat exchangers, and is finally expanded through an expansion
valve to the liquid receiver. The cold vapour from the liquid receiver is
returned through the heat exchangers to cool the incoming gas.
The yield and the specific work consumption may be computed by making the
mass and energy balance as in the Linde-Hampson system.
14.8 PRODUCTION OF SOLID ICE
Dry ice is used for low temperature refrigeration, such as to preserve ice cream and
other perishables. The property diagram of CO2 on the p-h coordinates is given in
Fig. 14.26. The schematic diagram of producing solid CO2 and the corresponding
p-h diagram are shown in Fig. 14.27 and Fig. 14.28 respectively.
587
Refrigeration Cycles
Solved Examples
Example 14.1 A cold storage is to be maintained at – 5°C while the surroundings are at 35°C. The heat leakage from the surroundings into the cold storage is
estimated to be 29 kW. The actual COP of the refrigeration plant used is one-third
that of an ideal plant working between the same temperatures. Find the power
required (in kW) to drive the plant.
T2
268
Solution COP (Ideal) =
=
= 6.7
308 - 268
T1 - T2
m
Make-up
(m f)
M
mf
13
mf
C
m
2
Cooling
water
3
me
me
3 WE
12
m
Purifier
5
11
Joule-Kelvin
refrigeration
system
10
me
Expander
4
m me
m me mf
9
6
7
8
mf
(a)
14
2
1
13
T
3
190 K
5
4
40
at
m
1a
tm
12
Ex
withpansi
loss on
es
150 K
11
6
h=
10
c
9
8
7
s
(b)
Fig. 14.25
Claude Cycle for Air Liquefaction
588
Engineering Thermodynamics
p
p cr = 73 atm
t cr = 31∞C
Liquid + Vapour
p ct = 5.11 atm
- 60∞C
Solid + Vapour
- 78.5∞C
Triple point line
1 atm
h
Fig. 14.26 p-h Diagram of CO2
70 atm
2
Condenser
3
Compressor
Expansion
valve
6
1
4
7
1 atm
Make-up
5 Separator
Solid CO 2
Fig. 14.27 Production of Dry Ice-flow Diagram
Q2
W
\ Power required to drive the plant (Fig. 14.29)
\
Actual COP = 1/3 ¥ 6.7 = 2.23 =
Surroundings
T 1 = 308 K
70 atm
Q1 = Q2 + W
Q1
2
p
3
R
W
Q2
1 atm
5
4
6
1
h
Fig. 14.28 Refrigeration Cycle of a
Dry Ice Plant on p-h Plot
7
Q 2 = 29 kW
Cold storage
T 2 = 268 K
Q 2 = 29 kW
Fig. 14.29
589
Refrigeration Cycles
W=
29
Q2
=
= 13 kW
2.23 2.23
Example 14.2 A refrigerator uses R-134a as the working fluid and operates on
an ideal vapour compression cycle between 0.14 MPa and 0.8 MPa. If the mass
flow rate of the refrigerant is 0.06 kg/s, determine (a) the rate of heat removal from
the refrigerated space, (b) the power input to the compressor, (c) the heat rejection
rate in the condenser, and (d) the COP.
Solution
are:
From the R-134a tables, the enthalpies at the four states (Fig. 14.30)
h1 = 236.04 kJ/kg
s1 = 0.9322 kJ/kg K = s2
For p2 = 0.8 MPa, s2 = 0.9322 kJ/kgK,
h2 = 272.05 kJ/kg, h3 = h4 = 93.42 kJ/kg
p
0.8 MPa
3
2
Q2 = 0.06 (236.04 – 93.42) = 8.56 kW
Wc = 0.06 (272.05 – 236.04) = 2.16 kW
Q1 = 0.06 (272.05 – 93.42) = 10.72 kW
Q
8.56
COP = 2 =
= 3.963
Wc 2.16
1
4
h
Fig. 14.30
Example 14.3 A simple R-12 plant is to develop 5 tonnes of refrigeration. The
condenser and evaporator temperatures are to be 40°C and –10°C respectively.
Determine (a) the refrigerant flow rate in kg/s, (b) the volume flow rate handled by
the compressor in m3/s, (c) the compressor discharge temperature, (d) the pressure
ratio, (e) the heat rejected to the condenser in kW, (f) the flash gas percentage after
throttling, (g) the COP, and (h) the power required to drive the compressor.
How does this COP compare with that of a Carnot refrigerator operating
between 40°C and – 10°C?
Solution From the table of the thermodynamic properties of a saturated
Refrigerant-12 (Fig. 14.31), p1 = (psat)–10°C = 2.1912 bar, h1 = 183.19 kJ/kg, s1 =
0.7019 kJ/kg K, v1 = 0.077 m3/kg, p2 = (psat)40°C = 9.6066 bar, and h3 = 74.59 kJ/kg
= h4. From the superheated table of R-12, when p2 = 9.6066 bar, and s2 = s1 =
0.7019 kJ/kg K, by interpolation, t2 = 48°C, and h2 = 209.41 kJ/kg. The capacity of
the plant = 5 ¥ 14,000 = 70,000 kJ/h. If w is the refrigerant flow rate in kg/s
70.000
w(h1 – h4) =
= 19.44 kW
3600
\
w=
19.44
= 0.18 kg/s
183.19 - 74.59
590
Engineering Thermodynamics
3
Wc
2
p
Q1
p2
4
1
40∞C
3
t2
s=
- 10∞C
p1
c
1
4
Q2
2
x4
h
(a)
(b)
Fig. 14.31
Volume flow rate = w ◊ v1 =0.18 ¥ 0.077 = 0.0139 m3/s
Compressor discharge temperature = 48°C
p
9.6066
Pressure ratio = 2 =
= 4.39
2.1912
p1
Heat rejected to the condenser = w(h2 – h3)
= 0.18 (209.41 – 74.59) = 24.27 kW
h4 = hf + x4hfg = 26.87 + x4 ¥ 156.31 = 74.59
\
\
47.72
= 0.305
156.31
Flash gas percentage = 30.5%
x4 =
COP =
h1 - h4
183.19 - 74.59
=
209. 41 - 183.19
h2 - h1
108.60
= 4.14
26.22
Power required to drive the compressor
= w(h2 – h1) = 0.18 ¥ 26.22 = 4.72 kW
=
COP(Reversible) =
\
T2
263
=
= 5.26
50
T1 - T2
COP (Vap.Comp.cycle) 4.14
=
= 0.787
5.26
COP (Carnot cycle)
Example 14.4 A Refrigerant-12 vapour compression plant producing 10 tonnes
of refrigeration operates with condensing and evaporating temperatures of 35°C
and – 10°C respectively. A suction line heat exchanger is used to subcool the saturated liquid leaving the condenser. Saturated vapour leaving the evaporator is
591
Refrigeration Cycles
superheated in the suction line heat exchanger to the extent that a discharge temperature of 60°C is obtained after isentropic compression. Determine (a) the
subcooling achieved in the heat exchanger, (b) the refrigerant flow rate in kg/s, (c)
the cylinder dimensions of the two-cylinder compressor, if the speed is 900 rpm,
stroke-to-bore ratio is 1.1, and the volumetric efficiency is 80%, (d) the COP of the
plant, and (e) the power required to drive the compressor in kW.
Solution From the p-h chart of R-12, the property values at the states, as shown
in Fig. 14.32.
h3 = 882, h2 = 1034
h6 = 998, h1 = 1008 kJ/kg
v1 = 0.084 m3/kg
h3 – h4 = h1 – h6
882 – h4 = 1008 – 998 = 10
\
h4 = 872 kJ/kg
\
t4 = 25°C
So 10°C subcooling is achieved in the heat exchanger. Refrigeration effect =
h6 – h5 = 998 – 872 = 126 kJ/kg
10 ¥ 14000
\
Refrigerant flow rate
= 1110 kg/h = 0.31 kg/s
126
Volume flow rate = w ◊ v1 = 1110 ¥ 0.084 = 93 m3/h
Compressor displacement =
Q1
93
= 116 m3/h = 1.94 m 3/min
0.8
Condenser
Wc
p
Suction
line heat
exchanger
4
35∞C
3
2
t2
- 10∞C 6
Compressor
5
Q2
s=
c
Expansion
valve
1
Evaporator
h
(a)
(b)
Fig. 14.32
592
Engineering Thermodynamics
p 2
D LNn
4
D = diameter
L = stroke
N = rpm
n = number of cylinders of the compressor.
p 2
D ¥ 1.1D ¥ 900 ¥ 2 = 1.94 m3/min
4
D3 = 1250 cm3
This is equal to
where
or
D = 10.8 cm
and
L = 11.88 cm
h6 - h5
126
=
= 4.85
1034 - 1008
h2 - h1
Power required to drive the compressor
1110 ¥ 26
= w(h2 – h1) =
= 8.02 kW
3600
COP =
Example 14.5 A two-stage vapour compression refrigeration system with a direct contact heat exchanger (flash chamber) operates with ammonia as the
refrigerant. The evaporator and condenser temperatures are – 30 and 40°C
respectively. If the capacity of the plant is 30 tonnes of refrigeration, estimate the
total work of compression and the COP. Had the compression been done in a single
stage, what would have been the percentage increase in the work of compression?
What is the percentage increase in the COP owing to the staging of the compression process?
Using tables for ammonia given in the appendix,
p
m1
5
p
Solution
4
7
1
8
2
2
3
6
40°C
3
m2
- 30°C
4
h
h
(a)
(b)
Fig. 14.33
at 40°C,
at – 30°C,
p2 = 1554.3 kPA
p1 = 119.5 kPa
\
pi = 1554.3 ¥ 119.5 = 431 kPa
1
Refrigeration Cycles
\
h1 = 1404.6
s2 = s1
h2 = 1574.3
h3 = 1443.5
s4 = s3
h4 = 1628.1
593
h5 = 371.7 = h6 , h7 = 181.5 hg
\
3.89 ¥ 30
116.7
=
= 0.0954 kg/s
1404.6 - 181.5 1223.1
1392.8
h -h
�1 = m
� 2 2 7 = 0.0954 ¥
m
= 0.124 kg/s
1071.8
h3 - h6
�2 =
m
� c (h2 – h1) + m
� 1(h4 – h3)
W� c = m
= 0.0954 ¥ 169.7 + 0.124 ¥ 184.6 = 16.19 + 22.89
= 39.08 kW
COP =
30 ¥ 3.89
= 2.986
39.08
Single stage
h1 = 1404.6
h2 = 1805.1
h3 = 371.7 = h4
� =
m
30 ¥ 3.89
116.7
=
= 0.113 kg/s
1404.6 - 371.7 1032 .9
� (h2 – h1) = 0.113 ¥ 400.5 = 45.26 kW
Wc = m
116.7
= 2.578
45.26
Increase in work of compression (for single stage)
45.26 - 39.08
=
¥ 100 = 15.81%
39.08
Increase in COP for 2-stage compression
\
COP =
=
2.986 - 2.578
¥ 100 = 15.82%
2.578
Example 14.6 In an aqua-ammonia absorption refrigerator system, heat is
supplied to the generator by condensing steam at 0.2 MPa, 90% quality. The
temperature to be maintained in the refrigerator is – 10°C, and the ambient
temperature is 30°C. Estimate the maximum COP of the refrigerator.
If the actual COP is 40% of the maximum COP and the refrigeration load is
20 tonnes, what will the required steam flow rate be?
At 0.2 MPa, from the steam table (Fig. 14.44)
tsat = 120.2°C, hfg = 2201.9 kJ/kg
The maximum COP of the absorption refrigeration system is given by Eq. (14.4)
(T - T ) T
(COP)max = 1 2 R
(T2 - TR ) T1
Solution
594
where
\
\
Engineering Thermodynamics
T1 = generator temperature
= 120.2 + 273 = 393.2 K
T2 = condenser and absorber temperature
= 30 + 273 = 303 K
TR = evaporator temperature
= – 10 + 273 = 263 K
(393.2 - 303) ¥ 263 90.2 ¥ 263
(COP)max =
=
= 1.5
(303 - 263) ¥ 393.2 40 ¥ 393.2
Actual COP = 1.5 ¥ 0.4 = 0.60
Since
COP =
QE
QG
20 ¥ 14000
QE
=
= 129.6 kW
COP 0.60 ¥ 3600
Heat transferred by 1 kg of steam on condensation
= (hf + xhfg) – hf = 0.9 ¥ 2201.9 = 1981.71 kJ/kg
\ Steam flow rate required
129.6
=
= 0.0654 kg/s
1981.71
QG =
NH 3 vapour
Steam
0.9 dry
QC
0.2 MPa
Generator
Cooling
water
Condenser
QG
Aqua-pump
Reducing
valve
Expansion valve
QE
NH 3 Vapour
QA
Absorber
Evaporator
Cooling water
Fig. 14.34
Example 14.7 In an aircraft cooling system, air enters the compressor at
0.1 MPa, 4°C, and is compressed to 0.3 MPa with an insentropic efficiency of 72%.
After being cooled to 55°C at constant pressure in a heat exchanger the air then
expands in a turbine to 0.1 MPa with an isentopic efficiency of 78%. The low
temperature air absorbs a cooling load of 3 tonnes of refrigeration at constant
595
Refrigeration Cycles
pressure before re-entering the compressor which is driven by the turbine. Assuming air to be an ideal gas, determine the COP of the refrigerator, the driving power
required, and the air mass flow rate.
Solution
Given: (Fig. 14.35)
2
2s
3
p
WC
4
1
4s
Q2
4
s
(a)
(b)
Fig. 14.35
T1 = 277 K, T3 = 273 + 55 = 328 K
T2 s
=
T1
\
FG p IJ
Hp K
( g -1)/ g
2
1
T2s = 277(3)0.4/1.4 = 379 K
T2s – T1 = 102 K
\
T2 – T1 =
T4 s
=
T3
\
102
= 141.8 K
0.72
FG p IJ
Hp K
( g -1)/ g
2
1
T4s = 328 (3)0.4/1.4 =
328
= 240 K
1.368
T3 – T4s = 88 K
\
1
1=
0.
1
C
M
T
Pa
p
2
2=
3
T
WE
0.3
MP
a
Q1
T3 – T4 = 0.78 ¥ 88 = 68.6 K
\
T4 = 259.4 K
Refrigerating effect = cp (T1 – T4) = 17.6 cp kJ/kg
Net work input = cp [(T2 – T1) – (T3 – T4)]
= cp [141.8 – 68.6] = 73.2 cp kJ/kg
COP =
17.6 c p
73.2 c p
= 0.24
4∞C
596
Engineering Thermodynamics
Driving power required
=
3 ¥ 14000
= 48.6 kW
0.24 ¥ 3600
Mass flow rate of air
3 ¥ 14000
= 2374.5 kg/h = 0.66 kg/s
1.005 ¥ 17.6
Example 14.8 A vapour-compression heat pump system uses R-12 as the working fluid. The refrigerant enters the compressor at 2.4 bar, 0°C with a volumetric
flow rate of 0.6 m3/min. Compression is adiabatic to 9 bar, 60°C and the saturated
liquid exits the condenser at 9 bar. Determine (a) the power input to the compressor, (b) the heating capacity of the system, (c) the coefficient of performance,
(d) the isentropic compressor efficiency.
=
Solution
b
At p1 = 2.4 bar, T1 = 0∞C
h1 = 188.99 kJ/kg
Adiabatic
s1 = 0.7177 kJ/kg K
p
v1 = 0.0703 m3 /kg
At p2 = 9 bar, T2 = 60°C
a
C
h2 = 219.37 kJ/kg
Isothermal
h2s = 213.27 kJ/kg
V
At p2 = 9 bar, sat. liquid
Fig. 14.36
h3 = 71.93 kJ/kg = h4
0.6 m3
1
1 1
m& = AV
kg m3
=
¥
60 s
0.0703
v1
= 0.1422 kg/s
(a) Power input W&c = m& (h2 – h1 )
= 0.1422 (219.37 – 188.99)
= 4.32 kW
Ans.
(b) Heating capacity, Q&1 = m& (h2 – h3)
= 0.1422 (219.37 – 71.93)
= 20.97 kW = 5.963 tonnes
Q&1 20.97
(c) COP = & =
= 4.854
4.32
Wc
Ans.
Ans.
597
Refrigeration Cycles
(d) h is =
h2 s - h1
213.27 - 188.99
=
= 0.799 or 79.9%
219.37 - 188.99
h2 - h1
Ans.
Example 14.9 In an air-refrigeration system working on reversed Brayton
cycle, the temperature of air at entrance to compressor (pressure ratio = 4, efficiency = 0.8) is 275 K and the inlet pressure is 1 bar. The pressure loss in the
cooler is 0.1 bar and in the cold chamber it is 0.08 bar. The temperature of air at
turbine (efficiency = 0.85) inlet is 310 K. Estimate the pressure ratio for the turbine and the COP of the cycle.
Solution
2s
T1 = 275 K, T3 = 310 K
2
p2
= 4, p1 = 1 bar
p1
T
p2
3
T2s = T1 (p2 /p1)
p1
= 275 ¥ 40.286
= 408.81 K
4s
1
4
T2 = T1 + (T2 s – T1) hc
= 442.26 K
Fig. 14.37
p3 = p2 – pr loss in the
cooler = 4 – 0.1 = 3.9 bar
p4 = p1 + pr loss in the cold chamber = 1 + 0.08
= 1.08 bar
\ Pressure ratio for the turbine =
3.9
= 3.611
1.08
Ans.
T3
= (p3/p4s)0.286 = (3.611)0.286 = 1.4437
T4 s
T4s = 214.72 K
T4 = T3 – (T3 – T4s)h t = 310 – (310 – 214.72) ¥ 0.85
= 229.015 K
\
COP =
275 - 229.015
T1 - T4
=
(442.26 - 310) - (275 - 229.015)
(T2 - T3 ) - (T1 - T4 )
= 0.533
Ans.
598
Engineering Thermodynamics
Review Questions
14.1 What is refrigeration? How is (a) ice and (b) dry ice used for the purpose of
refrigeration?
14.2 Explain the vapour compression cycle with the help of flow, T-s and p-h
diagrams. Can this cycle be reversible? If not, why?
14.3 What do you understand by dry and wet compression? Which is preferred and
why?
14.4 What is refrigerating effect?
14.5 What is a tonne of refrigeration?
14.6 Explan the effect of superheat and subcooling on the vapour compression cycle.
14.7 How does the actual vapour compression cycle differ from the ideal one?
14.8 What is a suction line heat exchanger? When is it used?
14.9 What are the expansion devices used in a vapour compression plant? When are
they used?
14.10 What are the different types of compressors used in vapour compression plants
and what are their applications?
14.11 What is a multistage vapour compression plant? When is it used?
14.12 What is a flash chamber? What is its advantage?
14.13 What are the most widely used refrigerants?
14.14 How are refrigerants numbered?
14.15 Why is the use of halogenated hydrocarbons as refrigerants now discouraged?
14.16 What are the effects of CFCs on the environment? How do they affect the ozone
layer?
14.17 What is ODP? Why has R-22 less ODP than R-12?
14.18 What are the parameters to be considered in the selection of a refrigerant?
14.19 What do you understand by cascade refrigeration system? Explain it with the
help of flow and T-s diagrams.
14.20 Evaluate ammonia as a refrigerant.
14.21 What is an absorption refrigeration cycle? How does it differ from a vapour
compression cycle?
14.22 How is the refrigerant liberated from the absorbent in an aqua-ammonia
system?
14.23 What are the functions of the analyzer and the rectifier?
14.24 What is the advantage of using a suction line heat exchanger in an absorption
refrigeration system?
14.25 In a lithium bromide-water absorption system, which is the refrigerant? What
is its limitation?
14.26 Derive the expression for the maximum COP of an absorption refrigeration
system.
14.27 How does a heat pump upgrade low grade reject heat?
14.28 How can a heat pump be used for (a) space heating (b) year-round
airconditioning?
14.29 How is a reversed Brayton cycle used for refrigeration?
14.30 Why is the COP of a gas cycle refrigeration system low?
14.31 Why is gas cycle refrigeration preferred in aircraft?
14.32 What is the principle of the Linde-Hampson system for liquefaction of air?
Refrigeration Cycles
599
14.33 Derive the expressions of liquid yield and the minimum work requirement in a
Linde-Hampson system.
14.34 How does Claude cycle differ from a Linde-Hampson cycle in the context of
the liquefaction of air?
14.35 With the help of flow and p-h diagrams, explain how dry ice is produced.
Problems
14.1 A refrigerator using R-134a operates on an ideal vapour compression cycle
between 0.12 and 0.7 MPa. The mass flow of refrigerant is 0.05 kg/s. Determine (a) the rate of heat removal from the refrigerated space, (b) the power
input to the compressor, (c) the heat rejection to the environment, and (d) the
COP.
Ans. (a) 7.35 kW, (b) 1.85 kW, (c) 9.20 kW, (d) 3.97
14.2 A Refrigerant-12 vapour compression cycle has a refrigeration load of
3 tonnes. The evaporator and condenser temperatures are – 20°C and 40°C
respectively. Find (a) the refrigerant flow rate in kg/s, (b) the volume flow rate
handled by the compressor in m3/s, (c) the work input to the compressor in
kW, (d) the heat rejected in the condenser in kW, and (e) the isentropic discharge temperature.
If there is 5 C deg. of superheating of vapour before it enters the compressor, and 5 C deg. subcooling of liquid before it flows through the expansion
valve, determine the above quantities.
14.3 A 5 tonne R-12 plant maintains a cold store at – 15°C. The refrigerant flow
rate is 0.133 kg/s. The vapour leaves the evaporator with 5 C deg. superheat.
Cooling water is available in plenty at 25°C. A suction line heat exchanger
subcools the refrigerant before throttling. Find (a) the compressor discharge
temperature, (b) the COP, (c) the amount of subcooling in C deg., and (d) the
cylinder dimensions of the compressor, if the speed is 900 rpm, stroke-to-bore
ratio is 1.2, and volumetric efficiency is 95%.
Allow approximately 5°C temperature difference in the evaporator and condenser.
Ans. (a) 66°C, (b) 4.1 (c) 125°C, (d) 104.5 mm, 125 mm
14.4 A vapour compression refrigeration system uses R-12 and operates between
pressure limits of 0.745 and 0.15 MPa. The vapour entering the compressor
has a temperature of – 10°C and the liquid leaving the condenser is at 28°C. A
refrigerating load of 2 kW is required. Determine the COP and the swept
volume of the compressor if it has a volumetric efficiency of 76% and runs at
600 rpm.
Ans. 4.15, 243 cm3
14.5 A food-freezing system requires 20 tonnes of refrigeration at an evaporator
temperature of – 35°C and a condenser temperature of 25°C. The refrigerant,
R-12, is subcooled 4°C before entering the expansion valve, and the vapour is
superheated 5°C before leaving the evaporator. A six-cylinder single-acting
compressor with stroke equal to bore is to be used, operating at 1500 rpm.
Determine (a) the refrigerating effect, (b) the refrigerant flow rate, (c) the theo-
600
14.6
14.7
14.8
14.9
14.10
14.11
14.12
14.13
14.14
Engineering Thermodynamics
retical piston displacement per sec, (d) the theoretical power required in kW,
(e) the COP, (f) the heat removed in the condenser, and (g) the bore and stroke
of the compressor.
A R-12 vapour compression refrigeration system is operating at a condenser
pressure of 9.6 bar and an evaporator pressure of 2.19 bar. Its refrigeration
capacity is 15 tonnes. The values of enthalpy at the inlet and outlet of the
evaporator are 64.6 and 195.7 kJ/kg. The specific volume at inlet to the reciprocating compressor is 0.082 m3/kg. The index of compression for the compressor is 1.13. Determine: (a) the power input in kW required for the compressor, and (b) the COP. Take 1 tonne of refrigeration as equivalent to heat
removal at the rate of 3.517 kW.
Ans. (a) 11.57 kW, (b) 4.56
A refrigeration plant produces 0.139 kg/s of the ice at – 5°C from water at
30°C. If the power required to drive the plant is 22 kW, determine the capacity
of the ice plant in tonnes and the actual COP (cp of ice = 2.1 kJ/kg K).
An air conditioning unit using R-12 as a refrigerant is mounted in the window
of a room. During steady operation 1.5 kW of heat is transferred from the air
in the room to the evaporator coils of R-12. If this air is at 22°C and the temperature of R-12 in the evaporator is 15°C, determine (a) the refrigerant flow
rate, and (b) the minimum power required to drive the compressor if the outside air is at 43°C and the temperature of the refrigerant during condensation
is 50°C.
In a solar energy operated aqua-ammonia absorption refrigeration system,
water is cooled at the rate of 10 kg/s from 38°C to 13°C. If the incident solar
energy is 640 W/m2 and the COP of the system is 0.32, estimate the area of the
solar collector needed.
A gas refrigerating system using air as a refrigerant is to work between –12°C
and 27°C using an ideal reversed Brayton cycle of pressure ratio 5 and minimum pressure 1 atm, and to maintain a load of 10 tonnes. Find (a) the COP,
(b) the air flow rate in kg/s, (c) the volume flow rate entering the compressor
in m3/s, and (d) the maximum and minimum temperatures of the cycle.
An open cycle (Brayton) aircraft cabin cooler expands air at 27°C through a
turbine which is 30% efficient from 2 to 1 atm. The cabin temperature is not to
exceed 24°C. Estimate the mass flow rate of air required (kg/s) for each tonne
of cooling.
Determine the ideal COP of an absorption refrigerating system in which the
heating, cooling, and refrigeration take place at 197°C, 17°C, and – 3°C respectively.
Ans. 5.16
A heat pump is to use an R-12 cycle to operate between outdoor air at – 1°C
and air in a domestic heating system at 40°C. The temperature difference in
the evaporator and the condenser is 8°C. The compressor efficiency is 80%,
and the compression begins with saturated vapour. The expansion begins with
saturated liquid. The combined efficiency of the motor and belt drive is 75%.
If the required heat supply to the warm air is 43.6 kW, what will be the electrical load in kW?
An ideal (Carnot) refrigeration system operates between the temperature limits of – 30°C and 25°C. Find the ideal COP and the power required from an
external source to absorb 3.89 kW at high temperature.
Refrigeration Cycles
601
14.15 A heat pump that operates on the ideal vapour compression cycle with R-134a
is used to heat a house and maintain it at 20°C, using underground water at
10°C as the heat source. The house is losing heat at a rate of 75 MJ/h. The
evaporator and condenser pressures are 320 and 800 kPa respectively. Determine the power input to the heat pump and the electric power saved by using a
heat pump instead of a resistance heater.
Ans. 2.27 kW, 18.56 kW
14.16 An ammonia-absorption system has an evaporator temperature of – 12°C and
a condenser temperature of 50°C. The generator temperature is 150°C. In this
cycle, 0.42 kJ is transferred to the ammonia in the evaporator for each kJ transferred to the ammonia solution in the generator from the high temperature
source.
It is desired to compare the performance of this cycle with the performance
of a similar vapour compression cycle. For this, it is assumed that a reservoir is
available at 150°C, and that heat is transferred from this reservoir to a reversible engine which rejects heat to the surroundings at 25°C. This work is then
used to drive an ideal vapour compression system with ammonia as the refrigerant. Compare the amount of refrigeration that can be achieved per kJ from
the high temperature source in this case with the 0.42 kJ that can be achieved
in the absorption system.
14.17 An R-12 plant is to cool milk from 30°C to 1°C involving a refrigeration capacity of 10 tonnes. Cooling water for the condenser is available at 25°C and
5 C deg. rise in its temperature is allowable. Determine the suitable condensing and evaporating temperatures, providing a minimum of 5 C deg. differential, and calculate the theoretical power required in kW and the cooling water
requirement in kg/s. Also, find the percentage of flash gas at the end of the
throttling. Assume a 2 C deg. subcooling in the liquid refrigerant leaving the
condenser.
14.18 The following data pertain to an air cycle refrigeration system for an aircraft.
Capacity 5 tonnes
Cabin air inlet temperature 15°C and outlet temperature 25°C
Pressure ratio across the compressor 5
The aircraft is flying at 0.278 km/s where the ambient conditions are 0°C and
80 kPa. Find the COP and the cooling effectiveness of the heat exchanger. The
cabin is at 0.1 MPa, and the cooling turbine powers the circulating fans.
14.19 A water cooler supplies chilled water at 7°C when water is supplied to it at
27°C at a rate of 0.7 litres/min., while the power consumed amounts to
200 watts. Compare the COP of this refrigeration plant with that of the ideal
refrigeration cycle for a similar situation.
14.20 A refrigerating plant of 8 tonnes capacity has an evaporation temperature of
– 8°C and condenser temperature of 30°C. The refrigerant, R-12, is subcooled
5°C before entering the expansion valve and the vapour is superheated 6°C
before leaving the evaporator coil. The compression of the refrigerant is
isentropic. If there is a suction pressure drop of 20 kPa through the valves, and
discharge pressure drop of 10 kPa through the valves, determine the COP of
the plant, theoretical piston displacement per sec. and the heat removal rate in
the condenser.
602
Engineering Thermodynamics
14.21. An ultra-low-temperature freezer system employs a coupling of two vapour
compression cycles of R-12 and R-13, as shown in Fig. 14.38. The states and
properties of both cycles are shown on the T-s plot. Determine the ratio of the
circulation rates of the two refrigerants, w1/w2 and the overall COP. How does
this COP compare with the Carnot COP operating between 42°C and – 70°C?
2
h 3 = 76
h 2 = 211
w1
3
b
42∞C
h b = 156
R = 12
T
C
h c = 44
4
O∞C
1
h 1 = 181
- 15∞C
R-13
w2
d
h a = 115
- 70∞C
a
s
Fig. 14.38
14.22 Derive an expression for the COP of an ideal gas refrigeration cycle with a
regenerative heat exchanger. Express the result in terms of the minimum gas
temperature during heat rejection (Th) maximum gas temperature during heat
absorption (T1), and pressure ratio for the cycle (p2/p1).
T1
Ans. COP =
(g -1)/ g
Th rp
- T1
14.23 Large quantities of electrical power can be transmitted with relatively little
loss when the transmission cable is cooled to a superconducting temperature.
A regenerated gas refrigeration cycle operating with helium is used to maintain an electrical cable at 15 K. If the pressure ratio is 10 and heat is rejected
directly to the atmosphere at 300 K, determine the COP and the performance
ratio with respect to the Carnot cycle.
Ans. 0.02, 0.38
14.24 A 100 tonne refrigerating plant using R-12 has a condensing temp. of 35°C
and an evaporating temp. of 5°C. Calculate the power requirement of the compressor in kW, the volume flow rate of compressor and the compressor displacement volume if the volumetric effy. is 75% and the mechanical efficiency
is 80%.
If a liquid suction heat exchanger is installed in the above plant which
subcools the condensed refrigerant to 30°C, what would be the refrigeration
capacity of the plant and the power required by the compressor?
14.25 A heat pump installation is proposed for a home heating unit with an output
rated at 30 kW. The evaporator temperature is 10°C and the condenser
Refrigeration Cycles
603
pressure is 0.5 bar. Using an ideal vapour compression cycle, estimate the
power required to drive the compressor if steam/water mixture is used as the
working fluid, the COP and the mass flow rate of the fluid. Assume saturated
vapour at compressor inlet and saturated liquid at condenser outlet.
Ans. 8.0 kW, 3.77, 0.001012 kg/s
14.26 A 100 tonne low temperature R-12 system is to operate on a 2-stage vapour
compression refrigeration cycle with a flash chamber, with the refrigerant
evaporating at – 40°C, an intermediate pressure of 2.1912 bar, and condensation at 30°C. Saturated vapour enters both the compressors and saturated liquid enters each expansion valve. Consider both stages of compression to be
isentropic. Determine: (a) The flow rate of refrigerant handled by each compressor, (b) the total power required to drive the compressor, (c) the piston
displacement of each compressor, if the clearance is 2.5% for each machine,
and (d) the COP of the system, (e) What would have been the refrigerant flow
rate, the total work of compression, the piston displacement in each compressor
and the compressor and the COP, if the compression had occurred in a single
stage?
Ans. (a) 2.464, 3.387 kg/s, (b) 123 kW, (c) 0.6274, 0.314 m3/s,
(d) 2.86, (e) 3.349 kg/s, 144.54 kW, 1.0236 m3/s, 2.433
14.27 A vapour compression plant uses R–134a as refrigerant. The evaporator
temperature is – 10°C. The condenser pressure is 7.675 bar and there is no
subcooling of the condensate. The flash chamber pressure is 4.139 bar and
the dry saturated vapour bled off from the flash chamber is mixed with the
refrigerant from the LP compressor before entering the HP compressor.
The liquid from the flash chamber is throttled before entering the evaporator. Assuming isentropic compression, calculate (a) COP of the plant
(b) mass flow of refrigerant in the evaporator when the power input to the
plant is 100 kW.
Ans. (a) 5.93, (b) 3.38 kg/s
14.28 The working fluid in a heat pump installation is ammonia. The ammonia
after evaporation to a dry saturated state at 2°C is compressed to a pressure
of 12.38 bar at which it is cooled and condensed to a saturated liquid state.
It then passes through a throttle valve and returns to the evaporator. Calculate the COP assuming that the isentropic efficiency of the compressor is
0.85.
Ans. 7.88
15
Psychrometrics
The properties of the mixtures of ideal gases were presented in Chapter 10. The
name ‘psychrometrics’ is given to the study of the properties of air-water vapour
mixtures. Atmospheric air is considered to be a mixture of dry air and water vapour.
The control of moisture (or water vapour) content in the atmosphere is essential for
the satisfactory operation of many processes involving hygroscopic materials like
paper and textiles, and is important in comfort air conditioning.
15.1 PROPERTIES OF ATMOSPHERIC AIR
Dry air is a mechanical mixture of the gases: oxygen, nitrogen, carbon dioxide,
hydrogen, argon, neon, krypton, helium, ozone, and xenon. However, oxygen and
nitrogen make up the major part of the combination. Dry air is considered to consist
of 21% oxygen and 79% nitrogen by volume, and 23% oxygen and 77% nitrogen
by mass. Completely dry air does not exist in nature.Water vapour in varying
amounts is diffused through it. If pa and pw are the partial pressures of dry air and
water vapour respectively, then by Dalton’s law of partial pressures
pa + pw = p
where p is the atmospheric pressure.
\ Mole-fraction of dry air, xa
=
pa
= pa
p
(∵ p = 1 atm.)
and mole reaction of water vapour, xw
=
pw
= pw
p
Since pw is very small, the saturation temperature of water vapour at pw is less
than atmospheric temperature, tatm (Fig. 15.1). So the water vapour in air exists in
the superheated state, and air is said to be unsaturated.
Relative humidity (R.H. f) is defined as the ratio of partial pressure of water
vapour, pw, in a mixture to the saturation pressure, ps, of pure water, at the same
temperature of the mixture (Fig. 15.1).
605
T
Psychrometrics
ps
tatm
Dew point
temperature td
pw
s
Fig. 15.1 States of Water Vapour in Mixture
R.H. (f) =
\
pw
ps
If water is injected into unsaturated air in a container, water will evaporate, which
will increase the moisture content of the air, and pw will increase. This will continue
till air becomes saturated at that temperature, and there will be no more evaporation
of water. For saturated air, the relative humidity is 100%. Assuming water vapour
as an ideal gas
pw V = mw RH 2O T = nw RT
and
ps V = ms RH2O T = ns RT
where V is the volume and T the temperature of air, the subscripts w and s indicating
the unsaturated and saturated states of air respectively.
f=
p w mw
=
ps
ms
mass of water vapour in a given
volume of air at temperature T
=
mass of water vapour when the same volume
of air is saturated at temperature T
=
nw x w
=
ns
xs
Specific humidity or humidity ratio, W, is defined as the mass of water vapour
(or moisture) per unit mass of dry air in a mixture of air and water vapour.
If
G = mass of dry air
m = mass of water vapour
W=
m
G
Specific humidity is the maximum when air is saturated at temperature T, or
Wmax = Ws =
ms
G
606
Engineering Thermodynamics
If dry air and water vapour behave as ideal gases
pwV = m Rw T
pa V = GRa T
p
R
p w 8.3143 / 28.96
m
= w ◊ a =
p a Rw
p - p w 8.3143 / 18
G
\
W=
\
W = 0.622
pw
p - pw
(15.1)
where p is the atmospheric pressure.
If pw is constant, W remains constant.
If air is saturated at temperature T
W = Ws = 0.622
ps
p - ps
where ps is the saturation pressure of water vapour at temperature T.
The degree of saturation, m, is the ratio of the actual specific humidity and the
saturated specific humidity, both at the same temperature T.
\
pw
p - pw
p
p - ps
W
m=
=
= w ◊
ps
Ws
ps p - pw
0.622
p - ps
If
f=
0.622
pw
= 0, pw = 0, xw = 0, W = 0, i.e. for dry air,
ps
m=0
If
f = 100%, pw = ps, W = Ws, m = 1
Therefore, m varies between 0 and 1.
If a mixture of air and superheated (or unsaturated) water vapour is cooled at
constant pressure, the partial pressure of each constituent remains constant until the
water vapour reaches the saturated state. Further cooling causes condensation. The
temperature at which water vapour starts condensing is called the dew point temperature, tdp, of the mixture (Fig. 15.1). It is equal to the saturation temperature at
the partial pressure, pw, of the water vapour in the mixture.
Dry bulb temperature (dbt) is the temperature recorded by the thermometer with
a dry bulb.
Wet bulb temperature (wbt) is the temperature recorded by a thermometer when
the bulb is enveloped by a cotton wick saturated with water. As the air stream flows
past it, some water evaporates, taking the latent heat from the water-soaked wick,
thus decreasing its temperature. Energy is then transferred to the wick from the air.
When equilibrium condition is reached, there is a balance between energy removed
from the water film by evaporation and energy supplied to the wick by heat transfer,
and the temperature recorded is the wet bulb temperature.
607
Psychrometrics
A psychrometer is an instrument which measures both the dry bulb and the wet
bulb temperatures of air. Figure 15.2 shows a continuous psychrometer with a fan
for drawing air over the thermometer bulbs. A sling psychrometer has the two thermometers mounted on a frame with a handle. The handle is rotated so that there is
good air motion. The wet bulb temperature is the lowest temperature recorded by
the moistened bulb.
Dry
bulb
Wet
bulb
Wick
Air flow
Fan
Water reservoir
Fig. 15.2 Dry and Wet Bulb Temperatures
At any dbt, the greater the depression (difference) of the wbt reading below the
dbt, the smaller is the amount of water vapour held in the mixture.
When unsaturated air flows over a long sheet of water (Fig. 15.3) in an insulated
chamber, the water evaporates, and the specific humidity of the air increases. Both
the air and water are cooled as evaporation takes places. The process continues
until the energy transferred from the air to the water is equal to the energy required
to vaporize the water. When this point is reached, thermal equilibrium exists with
respect to the water, air and water vapour, and consequently the air is saturated. The
equilibrium temperature is called the adiabatic saturation temperature or the thermodynamic wet bulb temperature. The make-up water is introduced at this temperature to make the water level constant.
2
Unsaturated air
Saturated air
G2, W2, = W2s
G1, m1, W1, t1, h1
m2, t2 = twb, h2 = h2s
Make-up
water (at t2)
Insulated chamber
Fig. 15.3
Water hf (at t2)
Adiabatic Saturation Process
The ‘adiabatic’ cooling process is shown in Fig. 15.4 for the vapour in the airvapour mixture. Although the total pressure of the mixture is constant, the partial
pressure of the vapour increases, and in the saturated state corresponds to the adiabatic saturation temperature. The vapour is initially at the dbt t db1 and is cooled
608
Engineering Thermodynamics
adiabatically to the dbt tdb2 which is equal to the adiabatic saturation temperature
twb2. The adiabatic saturation temperature and the wet bulb temperature are taken to
be equal for all practical purposes. The wbt lies between the dbt and dpt.
T
Adiabatic saturation process
tdb1
twb2
tdp1
1
saturated
air
(tdb2 = twb2)
p=c
Initially unsaturated air
2
Dew point of unsaturated
air at 1
s
Fig. 15.4 Adiabatic Cooling Process on T-s Plot
Since the system is insulated and no work is done, the first law yields
Gha1 + m1 hw1 + (m2 – m1 ) hf2 = Gha2 + m2 hw2
where (m2 – m1) is the mass of water added, hf 2 is the enthalpy of the liquid water at
t2 (= twb2), ha is the specific enthalpy of dry air, and hw is the specific enthalpy of
water vapour in air. Dividing by G, and since hw2 = hg2
ha1 + W1 hw1 + (W2 – W1) hf 2 = ha2 + W2 hg2
(15.2)
Solving for W1
W1 =
=
where
W2 =
bh - h g + W dh - h i
a2
a1
2
g2
f2
hw1 - h f 2
b
g
Cpa T2 - T1 + W2 ◊ h fg2
hw1 - h f 2
(15.3)
m2 ms
ps
=
= 0.622
p - ps
G
G
The enthalpy of the air-vapour mixture is given by
Gh = Gha + mhw
where h is the enthalpy of the mixture per kg of dry air (it is not the specific enthalpy of the mixture)
\
h = ha + Whw
(15.4)
15.2 PSYCHROMETRIC CHART
The psychrometric chart (Fig. 15.5) is a graphical plot with specific humidity and
partial pressure of water vapour as ordinates, and dry bulb temperature as
609
Psychrometrics
abscissa. The volume of the mixture (m3/kg dry air), wet bulb temperature, relative
humidity, and enthalpy of the mixture appear as parameters. Any two of these properties fix the condition of the mixture. The chart is plotted for one barometric pressure, usually 760 mm Hg.
f=
10
0%
Barometric pressure = 1 atm
Wet-bulb and
dew point
temperature
scales
Scale for the mixture
enthalpy per unit
mass of dry air
W
Twb
Twb
Twb
pw
%
f=
50
Volume
per unit
mass of
dry air
f = 10%
Dry-bulb Temperature, tdb (ºC)
pw (bar)
Vol. of
mixture
(m3/kg dry air)
h1
)
k
h(
g
J/k
f
y
dr
air
on
ati
e
lin
V
tur
0
=1
0%
Sa
1
h=c
v
h=c
f
wbt
Const. f lines
tdp1
twb1
tdb1
Dry bulb temperature, tdb (°C)
(b)
Fig. 15.5
Psychrometric Chart
Specific humidity, W (kg vap./kg dry air)
(a)
610
Engineering Thermodynamics
The constant wbt line represents the adiabatic saturation process. It also
coincides with the constant enthalpy line. To show this, let us consider the energy
balance for the adiabatic saturation process (Eq. 15.2).
ha1 + W1hw1 + (W2 – W1) hf2 = ha2 + W2hw2
Since ha + Whw = h kJ/kg dry air (Eq. 15.4)
h1 – W1hf 2 = h2 – W2hf 2
where subscript 2 refers to the saturation state, and subscript 1 denotes any state
along the adiabatic saturation path. Therefore
h – Whf 2 = constant
Since Whf 2 is small compared to h (of the order of 1 or 2%)
h = constant
indicating that the enthalpy of the mixture remains constant during an adiabatic
saturation process.
15.3
PSYCHROMETRIC PROCESS
(a) Sensible Heating or Cooling (at W = Constant) Only the dry bulb temperature of air changes. Let us consider sensible heating of air [Figs. 15.6(a), (b),
(c)] Balance of
Dry air
G1 = G2
1
2
m1
G1
h1
m2
G2
h2
Air in
Air out
W1
t1
W2
t2
Q1–2
W
(a)
h
T
h2
h1
t2
1
%
100
f =
pw
t1
1
2
f2
f1
tdp
t1
t2
DBT
s
(b)
(c)
Fig. 15.6
Sensible Heating
2
611
Psychrometrics
Moisture
m1 = m2
G 1 W1 = G 2 W 2
Energy
G1h1 + Q1–2 = G2 h2
Q1–2 = G(h2 – h1 )
or
Q1–2 = G [ha2 + W2 hw2 – (ha1 + W1 hw1)]
= G [1.005 (t 2 – t1) + W2 [hg2 + 1.88 (t2 – tdp2)]
– W1 [hg1 + 1.88 (t1 – tdp1)] kJ
where the cp of water vapour in a superheated state has been assumed to be
1.88 kJ/kg ºC.
Since
W1 = W2, hg2 = hg1
tdp2 = tdp1
Q1 – 2 = G [1.005 (t2 – t1) + 1.88 W (t2 – t1)]
= G (1.005 + 1.88 W) (t2 – t1)
Similar equations can be obtained for sensible cooling at constant humidity ratio.
(b) Cooling and Dehumidification When the humidity ratio of air decreases,
air is said to be dehumidified, and when it increases, air is humidified. Air may be
cooled and dehumidified (a) by placing the evaporator coil across the air flow
(Fig. 15.7b), (b) by circulating chilled water (or brine) in a tube placed across the
air flow (Fig. 15.7c), or (c) by spraying chilled water to air in the form of fine mist
(Fig. 15.7d) to expose a large surface area. The temperature of the cooling surface
or the spray water must be below the dew point at state 1 (Fig. 15.7a). If the cooling
surface or the spray shower is of large magnitude, the air may come out at the
saturation state, 2s, known as the apparatus dew point (adp) (Fig. 15.8).
G1 = G2 = G
h1
W
f1
h2
h
1
W1
W2
f2
tdp2
twb2
2
tdp1 twb1
DBT
(a)
612
Engineering Thermodynamics
Evaporator
1
2
G2
m2
W2
Air out
G1
m1
W1
Air in
h1
t1
h2
t2
Q1–2
Expansion valve
Compressor
Condenser
(b)
1
2
Chilled
water
Air out
Air in
Evaporator
Expansion valve
Water
Compressor
R-22
Condenser
(c)
1
Spray shower
of chilled
water
2
Water eliminator
Air out
Air in
Evaporator
Make-up water
Pump
Expansion
valve
Chilled water
Compressor
R-22
(d)
Condenser
Fig. 15.7 Cooling and Dehumidification
613
h1
h3
LHL
h2
W
Psychrometrics
1
h
Sat.
line
2
2s
a.d.p.
3
SHL
tdp1
DBT
Fig. 15.8 Sensible and Latent Heat Loads
If L is the amount of water vapour removed
m1 = m2 + L
or
L = G (W1 – W2)
The energy equation gives
G1h1 = G2h2 + Q1–2 + L ◊ hf2
where hf 2 is the specific enthalpy of water at temperature t2 .
\
Q1–2 = G[(h1 – h2) – (W1 – W2)hf 2]
If (W1 – W2)hf 2 is small, the amount of heat removed is
Q1–2 = G(h1 – h2) = Total heat load (THL) on the cooling coil (kJ/h).
Now, as shown in Fig. 15.8, h1 – h2 = (h1 – h3) + (h3 – h2) = LHL + SHL = THL
where h1 – h3 = enthalpy change at constant dbt, known as latent heat load (LHL),
and (h3 – h2) is the enthalpy change at constant W, known as sensible heat load
(SHL).
Cooling and dehumidification of air is common in summer air conditioning.
(c) Heating and Humidification
The addition of heat and moisture to air is a
problem for winter air conditioning. Figure 15.9 shows this process. The water
added may be liquid or vapour. The following equations apply
G1 = G2 = G
m1 + L = m2
\
L = m2 – m1 = G(W2 – W1)
G1h1 + L ◊ hf + Q1 – 2 = G2h2
Q1 – 2 = G(h2 – h1) – G(W2 – W1)hf
= G[(h2 – h1 ) – (W2 – W1)hf]
Figure 15.10 shows first dehumidification and then heating.
(d) Adiabatic Mixing of Two Streams This is a common problem in air conditioning, where ventilation air and some room air are mixed prior to processing it
to the desired state (say, by cooling and dehumidification), and supplying it to the
614
Engineering Thermodynamics
2
G1
m1
Air in
h1
W1
t1
G2
m2
Air out
h2
W2
t2
L
hf
Water
Heating coil
(a)
h2
W
h1
W2
h
2
f=
%
100
W1
1
t wb1 t 1
t wb2 t 2
DBT
(b)
Fig. 15.9
1
Heating and Humidification
Heating
coil
Cooling
coil
mr
i
e 2
3
f2 = 100%
T2 < T1
w2 < w1
Moist air
m a,
T 1, w1
p = 1 atm
T3 > T2
w3 = w2
Initial dew
point
f1
1
f
mw
(Heating section)
Condensate – saturated at T 2
(Dehumidifier section)
(a)
00%
=1
f3
3
2
T2
T3
T1
Dry-bulb temperature
(b)
Fig. 15.10 Dehumidification with heating. (a) Equipment Schematic
(b) Psychrometric Chart Representation
w
615
Psychrometrics
conditioned space. The process is shown in Fig. 15.11. The following equations
hold good
G1 + G2 = G3
G1W1 + G2W2 = G3W3
G1h1 + G2h2 = G3h3
Combining these equations and rearranging
G1
h - h2 W3 - W2
= 3
=
h1 - h3 W1 - W3
G2
The points 1, 2 and 3 fall in a straight line, and the division of the line is inversely proportional to the ratio of the mass flow rates.
G1
W
h1 1
h1
1
h3
G3
h3
W3
2
h
G1
1
W1
W
W3
3
W2
2
3
G2
h2 2
W
G2
h2
t2 t3
t1
DBT
(a)
(b)
Fig. 15.11
Adiabatic Mixing of Two Air Streams
(e) Chemical Dehumidification Some substances like silica gel (product of
fused sodium silicate and sulphuric acid), and activated alumina have great affinity
with water vapour. They are called adsorbents. When air passes through a bed of
silica gel, water vapour molecules get adsorbed on its surface. Latent heat of condensation is released. So the dbt of air increases. The process is shown in
Fig. 15.12.
(f) Adiabatic Evaporative Cooling
A large quantity of water is constantly
circulated through a spray chamber. The air-vapour mixture is passed through the
spray and, in doing so, evaporates some of the circulating water. The air may leave
at a certain humidity ratio or in a saturated state (Fig. 15.13). The increase in specific humidity is equal to the quantity of water evaporated per unit mass of dry air.
No heat transfer takes place between the chamber and the surroundings. Therefore,
the energy required for evaporation is supplied by the air, and consequently, the
DBT is lowered. After the process has been in operation for a sufficient length of
time, the circulating water approaches the wbt of air.
616
Engineering Thermodynamics
i
t. l
Sa
ne
W1
1
2
DBT
t1
W
W2
t2
Fig. 15.12 Chemical Dehumidification
1
2s
h
W2s
W2
2
W
h 1 = h2
Insulation
Spray
G2
m2
h1
W1
t1
h2
W2
t2
L
W1
1
2
G1
m1
Water hf
twb t2
t1
DBT
(a)
(b)
Fig. 15.13
Adiabatic Evaporative Cooling
G1 = G2 = G
G1W1 + L = G2W2
\
L = G(W2 – W1)
G1h1 + Lhf = G2 ◊ h2
\
G(h1 – h2 ) + G(W2 – W1)hf = 0
\
h1 – W1hf = h2 – W2hf
The cooling tower utilizes the phenomenon of evaporative cooling to cool warm
water below the dbt of the air. However, the water never reaches the minimum
temperature, i.e. the wbt, since an excessively large cooling tower would then be
required. Also, since warm water is continuously introduced to the tower
(Fig. 15.14), the equilibrium conditions are not achieved, and the DBT of the air is
increased. Hence, while the water is cooled, the air is heated and humidified.
617
Psychrometrics
Air out
x, Make-up water
2
3
Warm
water (mw)
Air in
1
Cooled water
4
Fig. 15.14 Cooling Tower
The warm water is introduced at the top of the tower in the form of spray to
expose a large surface area for evaporation to take place. The more the water evaporates, the more is the effect of cooling. Air leaves from the top very nearly saturated. The following equations apply
G1 = G2 = G
G1W1 + mw3 = G2W2 + mw4
\
mw3 – mw4 = G(W2 – W1)
G1h1 + mw3 hw3 = G2h2 + mw4 hw4
\
G (h1 – h2) + mw3 hw3 = mw4 hw4
The difference in temperature of the cooled-water temperature and the wet
bulb temperature of the entering air is known as the approach. The range is the
temperature difference between the inlet and exit states of water. Cooling towers
are rated in terms of approach and range.
If x is the make-up water supplied, then (Fig. 15.14)
x = G(W2 – W1)
and
mw3 = mw4 = mw
By energy balance,
G1h1 + mw3 hw2 + xhw3 = G2h2 + mw3 hw4
\
mw(hw3 – hw4) = G(h2 – h1) – G(W2 – W1)hw
G
\
hw3 – hw4 =
[(h2 – h1) – (W2 – W1)hw]
mw
\
Range = tw3 – tw4 =
G
[(h2 – h1) – (W2 – W1)hw]
mw c pw
(15.5)
(15.6)
where cpw is the specific heat of water and hw is the enthalpy of make-up water.
Approach = tw3 – twb1
(15.7)
618
Engineering Thermodynamics
The active portion of the tower in which energy exchange occurs is filled with a
packing which breaks up the flow and exposes large surface area of water in contact
with the air.
The schematic layout of a cooling tower is shown in Fig. 15.15.
Discharged moist air
ma, T, w2 > w1
2
Fan
Warm water inlet
T3, mw
3
Atmospheric air
1
ma, T3, w3
4 Return water
mw
T4 < T 3
Liquid
5 Make-up
water
Fig. 15.15 Schematic of a Cooling Tower
Solved Examples
Example 15.1 Atmospheric air at 1.0132 bar has a dbt of 32ºC and a wbt of
26ºC. Compute (a) the partial pressure of water vapour, (b) the specific humidity,
(c) the dew point temperature, (d) the relative humidity, (e) the degree of saturation,
(f) the density of the air in the mixture, (g) the density of the vapour in the mixture,
and (h) the enthalpy of the mixture.
Solution The state of air is shown on the DBT-W plot in Fig. 15.16. The path
1–2 represents the constant wbt and enthalpy of the air, which also holds good
approximately for an adiabatic saturation process. From Eq. (15.1), the specific
humidity at state 2 is given by
W2 = 0.622
ps
p - ps
619
Psychrometrics
W
h1
2
h
1
W1
f1
26∞C 32∞C
DBT
Fig. 15.16
The saturation pressure ps at the wbt of 26ºC is 0.03363 bar.
\
W2 = 0.622
0.03363
= 0.021148 kg vap./kg dry air
1.0132 - 0.03363
From Eq. (15.3), for adiabatic saturation
W1 =
b
g
c pa T2 - T1 + W2 ◊ h fg2
hw1 - h f 2
From the steam tables, at 26ºC
hfg2 = 2439.9 kJ/kg, hf 2 = 109.1 kJ/kg
At
(a)
32ºC, hw1 = hg = 2559.9 kJ/kg
W1 =
a
f
a
= 0.0186 kg vap./kg dry air
(b)
W1 = 0.622
pw
= 0.0186
p - pw
p - pw
0.622
=
= 33.44
0.0186
pw
pw = 0.03 bar
(c)
(d)
Saturation temperature at 0.03 bar, dpt = 24.1ºC
Relative humidity, f =
f
1.005 25 - 32 + 0.021148 2439.9
2559.9 - 109.1
pw
psat
At 32ºC, psat = 0.048 bar
620
f=
\
(e)
Engineering Thermodynamics
Deg. of saturation m =
0.03
= 0.625 or 62.5%
0.048
a
f
f
W
p p - ps 0.03 1.0132 - 0.048
= w
=
Ws
ps p - pw 0.048 1.0132 - 0.03
a
= 0.614
(f)
Partial pressure of dry air
pa = p – pw = 1.0132 – 0.03 = 0.9832 bar
\ Density of dry air
ra =
(g)
pa
0.9832 ¥ 100
=
= 1.12 kg/m3 dry air
Ra Tdb 0.287 ¥ 273 + 32
a
f
Density of water vapour
rw = 0.0186
kg vap.
kg dry air
¥ 1.12 3
kg dry air
m dry air
= 0.021 kg vap/m3 dry air
(h)
Enthalpy of the mixture
h = ha + Whw = cpt a + W[hg + 1.88 (tdb – t dp)]
= 1.005 ¥ 32 + 0.0186 [2559.9 + 1.88 (32 – 24.1)]
= 80.55 kJ/kg
Example 15.2 An air-water vapour mixture enters an adiabatic saturator at
30ºC and leaves at 20ºC, which is the adiabatic saturation temperature. The pressure remains constant at 100 kPa. Determine the relative humidity and the humidity ratio of the inlet mixture.
Solution The specific humidity at the exit
ps
p - ps
W2 = 0.622
= 0.622
FG 2.339 IJ = 0.0149 kg vap.
kg dry air
H 100 - 2.339 K
The specific humidity at the inlet (Eq. 15.3)
W1 =
=
b
g
C pa T2 - T1 + W2 h fg2
hw1 - h f 2
a
f
1.005 20 - 30 + 0.0149 ¥ 2454.1
2556.3 - 83.96
= 0.0107 kg vap./kg dry air
621
Psychrometrics
W1 = 0.622
FG p IJ = 0.0107
H 100 - p K
w1
w1
pw1 = 1.691 kPa
f1 =
\
pw1 1.691
=
= 0.398 or 39.8%
ps1 4.246
Example 15.3 Saturated air at 2ºC is required to be supplied to a room where
the temperature must be held at 20ºC with a relative humidity of 50%. The air is
heated and then water at 10ºC is sprayed in to give the required humidity.
Determine the temperature to which the air must be heated and the mass of spray
water required per m3 of air at room conditions. Assume that the total pressure is
constant at 1.013 bar and neglect the fan power.
Solution The process is shown in Fig. 15.17. From the steam tables, at 20ºC, psat
= 2.339 kPa
H
50%
%
R
W
Q2
1
10
3
0
2
20∞C, 50% RH
Room
Heater
Sat. air
at 2∞C
2∞C
20∞C
Spary
DBT
(a)
(b)
Fig. 15.17
f3 =
pw3
p
= w3 = 0.50
psat t 3 2.339
b g
\
pw3 = 1.17 kPa
\
pa3 = 101.3 – 1.17 = 100.13 kPa
117
.
p
W3 = 0.622 w3 = 0.622 ¥
= 0.00727
100.13
pa 3
f1 =
pw1
bp g
= 1.00
sat 2 º C
At
2ºC, psat = 0.7156 kPa
\
pw1 = 0.7156 kPa
pa1 = 101.3 –0.7156 = 100.5844 kPa
622
Engineering Thermodynamics
W1 = 0.622
0.7156
= 0.00442
100.5844
W3 – W1 = 0.00727 – 0.00442 = 0.00285 kg vap./kg dry air
va3 =
\
Ra T3 0.287 ¥ 293
=
= 0.84 m 3/kg dry air
pa 3
100.13
Spray water = 0.00285
kg vap.
kg dry air
¥
kg dry air
0.84 m 3
= 0.00339 kg moisture/m3
G2h2 + mw4 h4 = G3h3
\
h2 + (W3 – W2)h4 = h3
ha2 + W2hw2 + (W3 – W2)h4 = ha3 + W3hw3
\
cp(t3 – t2) + W3hw3 – W2hw2 – (W3 – W2)h4 = 0
From the steam tables, at pw = 1.17 kPa
hg = 2518 kJ/kg and tsat = 9.65ºC
1.005 (20 – t2) + 0.00727 [2518 + 1.884 (20 – 9.65)]
– 0.00442 [2518 + 1.884 (t2 – 9.65)]
– 0.00285 ¥ 10 = 0, \
t2 = 27.2ºC
Example 15.4 An air conditioning system is designed under the following
conditions:
Outdoor conditions—30ºC dbt, 75% R.H.
Required indoor conditions—22ºC dbt, 70% R.H.
Amount of free air circulated—3.33 m3/s
Coil dew point temperature—14ºC
The required condition is achieved first by cooling and dehumidification, and
then by heating. Estimate (a) the capacity of the cooling coil in tonnes, (b) the
capacity of the heating coil in kW, and (c) the amount of water vapour removed in
kg/s.
Solution The processes are shown in Fig. 15.18. The property values, taken from
the psychrometric chart, are
h1 = 82, h2 = 52, h3 = 47, h4 = 40 kJ/kg dry air
W1 = 0.020, W2 = W3 = 0.0115 kg vap./kg dry air
v1 = 0.887 m3/kg dry air
3.33
= 3.754 kg dry air/sec
0.887
Cooling coil capacity = G(h1 – h3) = 3.754 (82 – 47) kJ/s
G=
\
= 3.754 ¥ 35 ¥ 3600 = 33.79 tonnes
14,000
623
Psychrometrics
Capacity of the heating coil = G(h2 – h3) = 3.754 (52 – 47) kJ/s
= 3.754 ¥ 5 = 18.77 kW
Heating coil
1
3
2
Moisture removed
Cooling coil
(a)
W
h1
h2
75%
1
h3
h4
2
3
h
4
W1
W2
70%
14∞C
22∞C 30∞C
DBT
(b)
Fig. 15.18
Rate of water vapour removed = G (W1 – W3 )
= 3.754 ¥ (0.0200 – 0.0115)
= 0.0319 kg/s
Example 15.5 Air at 20ºC, 40% RH is mixed adiabatically with air at 40ºC,
40% RH in the ratio of 1 kg of the former with 2 kg of the latter (on dry basis). Find
the final condition of air.
Solution.
tions
Figure 15.19 shows the mixing process of two air streams. The equaG1 + G2 = G3
G1W1 + G2W2 = G3W3
G1h1 + G2h2 = G3h3
result in
W2 - W3
h -h
G
= 2 3 = 1
W3 - W1
h3 - h1 G2
624
Engineering Thermodynamics
W
h2
h3
f
2
%
1
0
G1
h1
3
G3
f=
1
W2
3
0
G1
W3
h
1
G2
W1
f = 40%
G2
2
t3
20∞C
40∞C
DBT
(a)
(b)
Fig. 15.19
From the psychrometric chart
W1 = 0.0058, W2 = 0.0187 kg vap./kg dry air
h1 = 35, h2 = 90 kJ/kg dry air
1
0.0187 - W3
G
= 1 =
W3 - 0.0058
G2 2
2
1
W3 = ¥ 0.187 + ¥ 0.0058
3
3
= 0.0144 kg vap./kg dry air
\
\
Again
h2 - h3
G
1
= 1 =
h3 - h1
G2 2
\
h3 =
2
1
2
1
h2 + h1 = ¥ 90 + ¥ 35
3
3
3
3
= 71.67 kJ/kg dry air
\
Final condition of air is given by
W3 = 0.0144 kg vap./kg dry air
h3 = 71.67 kJ/kg dry air
Example 15.6 Saturated air at 21ºC is passed through a drier so that its final
relative humidity is 20%. The drier uses silica gel adsorbent. The air is then passed
through a cooler until its final temperature is 21ºC without a change in specific
humidity. Find out (a) the temperature of air at the end of the drying process, (b)
the heat rejected in kJ/kg dry air during the cooling process, (c) the relative humidity at the end of the cooling process, (d) the dew point temperature at the end of
625
Psychrometrics
the drying process, and (e) the moisture removed during the drying process in kg
vap./kg dry air.
Solution
From the psychrometric chart (Fig. 15.20)
T2 = 38.5ºC
h1 – h3 = 60.5 – 42.0 = 18.5 kJ/kg dry air
f3 = 53% t4 = 11.2ºC
W1 – W2 = 0.0153 – 0.0083 = 0.0070 kg vap/kg dry air
h1
h3
1
3
2
%
4
h
W1
W2
f
2=
20
f3
W
11.2ºC
21ºC
38.5∞C
DBT
Fig. 15.20
Example 15.7 For a hall to be air-conditioned, the following conditions are
given
Outdoor condition—40ºC dbt, 20ºC wbt
Required comfort condition—20ºC dbt, 60% RH
Seating capacity of hall—1500
Amount of outdoor air supplied—0.3 m3/min per person
If the required condition is achieved first by adiabatic humidification and then
by cooling, estimate (a) the capacity of the cooling coil in tonnes, and (b) the
capacity of the humidifier in kg/h.
Solution
From the psychrometric chart (Fig. 15.21)
h1 = h2 = 57.0, h3 = 42.0 kJ/kg dry air
W1 = 0.0065, W2 = W3 = 0.0088 kg vap./kg dry air
t2 = 34.5ºC, v1 = 0.896 m3/kg dry air
Amount of dry air supplied
G=
\
1500 ¥ 0.3
= 502 kg/min
0.896
Capacity of the cooling coil
= G(h2 – h3) = 502 (57 – 42) kJ/min
626
10
0%
Engineering Thermodynamics
W
=
h1
h
f
h3
3
2
f = 60%
20∞C
1
W2
W1
40∞C
DBT
Fig. 15.21
=
502 ¥ 15 ¥ 60
= 32.27 tonnes
14,000
Capacity of the humidifier
= G (W2 – W1) = 502 (0.0088 – 0.0065) kg/min
= 502 ¥ 23 ¥ 10–4 ¥ 60 = 69.3 kg/h
Example 15.8 Water at 30ºC flows into a cooling tower at the rate of 1.15 kg
per kg of air. Air enters the tower at a dbt of 20ºC and a relative humidity of 60%
and leaves it at a dbt of 28ºC and 90% relative humidity. Make-up water is supplied at 20ºC. Determine: (i) the temperature of water leaving the tower, (ii) the
fraction of water evaporated, and (iii) approach and range of the cooling tower.
Solution Properties of air entering and leaving the tower (Fig. 15.13) are
twb1 = 15.2ºC
twb2 = 26.7ºC
h1 = 43 kJ/kg dry air
h2 = 83.5 kJ/kg dry air
W1 = 0.0088 kg water vapour/kg dry air
W2 = 0.0213 kg water vapour/kg dry air
Enthalpies of the water entering the tower and the make-up water are
hw3 = 125.8 kJ/kg
hm = 84 kJ/kg
From the energy balance Eq. (15.5),
hw3 – hw4 =
=
G
[(h2 – h1) – (W2 – W1) hw]
mw
1
[(83.5 – 43) – (0.0213 – 0.0088) 84]
115
.
627
Psychrometrics
= 34.2 kJ/kg
Temperature drop of water
tw3 – tw4 =
tw4 = 21.8ºC
\
\
34.2
= 30 – tw4
4.19
Approach = tw4 – twb1 = 21.8 – 15.2 = 6.6ºC
Range = tw3 – tw4 = 30 – 21.8 = 8.2ºC
Fraction of water evaporated, x = G(W2 – W1) = 1(0.0213 – 0.0088)
= 0.0125 kg/kg dry air
Example 15.9 Water from a cooling system is itself to be cooled in the cooling
tower at a rate of 2.78 kg/s. The water enters the tower at 65ºC and leaves a collecting tank at the base at 30ºC. Air flows through the tower, entering the base at
15ºC, 0.1 MPa, 55% RH, and leaving the top at 35ºC, 0.1 MPa, saturated. Makeup water enters the collecting tank at 14ºC. Determine the air flow rate into the
tower in m3 /s and the make-up water flow rate in kg/s.
Solution Figure 15.22 shows the flow diagram of the cooling tower. From the
steam tables
at 15ºC, psat = 0.01705 bar, hg = 2528.9 kJ/kg
at 35ºC, psat = 0.05628 bar, hg = 2565.3 kJ/kg
pw
f1 =
= 0.55
psat 15ºC
b g
Hot saturated air (35∞C, 100% RH)
2
Hot water (65∞C, m w)
3
5
1
Make-up
water (14∞C)
Air
(15∞C, 55% RH)
4
Fig. 15.22
Cold water
(30∞C, m w)
628
Engineering Thermodynamics
pw1 = 0.55 ¥ 0.01705 = 0.938 ¥ 10–2 bar
pw
f2 =
= 1.00
psat 35ºC
\
b g
pw2 = 0.05628 bar
\
W1 = 0.622
0.938 ¥ 10 -2
pw
= 0.622 ¥
1.00 - 0.00938
p - pw
= 0.00589 kg vap./kg dry air
W2 = 0.622 ¥
\
0.05628
= 0.0371 kg vap./dry air
1.00 - 0.05628
Make-up water = W2 – W1 = 0.0371 – 0.00589
= 0.03121 kg vap./kg dry air
Energy balance gives
H2 + H4 – H1 – H3 – H5 = 0
For 1 kg of dry air
.
cpa (t2 – t1) + W2 h2 – W1 h1 + m w (h4 – h3) – (W2 – W1 ) h5 = 0
\
1.005 (35 – 15) + 0.0371 ¥ 2565.3 – 0.00589 ¥ 2528.9
.
+ m w (– 35) 4.187 – 0.03121 ¥ 4.187 ¥ 14 = 0
.
or
146.55 m w = 98.54
\
.
m w = 0.672 kg water/kg dry air
Since water flow rate is 2.78 kg/s
\
Rate of dry air flow =
\
Make-up water flow rate
2.78
= 4.137 kg/s
0.672
= 0.03121 ¥ 4.137 = 0.129 kg/s
Rate of dry air flow = 4.137 kg/s
Rate of wet air flow = 4.137 (1 + W1)
= 4.137 ¥ 1.00589 = 4.16 kg/s
\ Volume flow rate of air
=
.
4.16 ¥ 0.287 ¥ 288
m a RT
= 3.438 m3/s
=
100
p
Review